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How to determine the proton concentration in the following solution in mol/L?
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A $pu0.8L$ solution contains $pu0.60 g ce NaOH$ in water. Then the above question follows.
It's been a long time that I made some chemistry exercises. Now I'm stuck with a couple of hesitations.
I know one needs to calculate the number of protons of the molecule sodium hydroxide. But do you need to take water into account or not? I'm thinking not but then the question asks for the solution. So I probably should do that which gets me to $pu29 g mol-1$. Otherwise it is $pu20 g mol-1$.
So that I would divide by $pu0.60g$ which gives me for the first one $pu48 mol$ and the second one $pu33 mol$. Then I'd divide it by $pu8.0L$.
I probably should have a larger answer than what I've done so it is not correct but I don't know exactly how I should work this out.
I'd appreciate a hint.
solutions concentration
edited 12 mins ago
A.K.
6,55221247
asked 2 hours ago
Anonymous I
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Anonymous I is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
1
down vote
favorite
A $pu0.8L$ solution contains $pu0.60 g ce NaOH$ in water. Then the above question follows.
It's been a long time that I made some chemistry exercises. Now I'm stuck with a couple of hesitations.
I know one needs to calculate the number of protons of the molecule sodium hydroxide. But do you need to take water into account or not? I'm thinking not but then the question asks for the solution. So I probably should do that which gets me to $pu29 g mol-1$. Otherwise it is $pu20 g mol-1$.
So that I would divide by $pu0.60g$ which gives me for the first one $pu48 mol$ and the second one $pu33 mol$. Then I'd divide it by $pu8.0L$.
I probably should have a larger answer than what I've done so it is not correct but I don't know exactly how I should work this out.
I'd appreciate a hint.
solutions concentration
edited 12 mins ago
A.K.
6,55221247
asked 2 hours ago
Anonymous I
1115
New contributor
Anonymous I is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
@Karl: I meant number of protons in the molecule natriumhydroxide. I changed it.
– Anonymous I
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A $pu0.8L$ solution contains $pu0.60 g ce NaOH$ in water. Then the above question follows.
It's been a long time that I made some chemistry exercises. Now I'm stuck with a couple of hesitations.
I know one needs to calculate the number of protons of the molecule sodium hydroxide. But do you need to take water into account or not? I'm thinking not but then the question asks for the solution. So I probably should do that which gets me to $pu29 g mol-1$. Otherwise it is $pu20 g mol-1$.
So that I would divide by $pu0.60g$ which gives me for the first one $pu48 mol$ and the second one $pu33 mol$. Then I'd divide it by $pu8.0L$.
I probably should have a larger answer than what I've done so it is not correct but I don't know exactly how I should work this out.
I'd appreciate a hint.
solutions concentration
edited 12 mins ago
A.K.
6,55221247
asked 2 hours ago
Anonymous I
1115
New contributor
Anonymous I is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
A $pu0.8L$ solution contains $pu0.60 g ce NaOH$ in water. Then the above question follows.
It's been a long time that I made some chemistry exercises. Now I'm stuck with a couple of hesitations.
I know one needs to calculate the number of protons of the molecule sodium hydroxide. But do you need to take water into account or not? I'm thinking not but then the question asks for the solution. So I probably should do that which gets me to $pu29 g mol-1$. Otherwise it is $pu20 g mol-1$.
So that I would divide by $pu0.60g$ which gives me for the first one $pu48 mol$ and the second one $pu33 mol$. Then I'd divide it by $pu8.0L$.
I probably should have a larger answer than what I've done so it is not correct but I don't know exactly how I should work this out.
I'd appreciate a hint.
solutions concentration
solutions concentration
edited 12 mins ago
A.K.
6,55221247
asked 2 hours ago
Anonymous I
1115
New contributor
Anonymous I is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 12 mins ago
A.K.
6,55221247
asked 2 hours ago
Anonymous I
1115
New contributor
Anonymous I is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 12 mins ago
A.K.
6,55221247
edited 12 mins ago
A.K.
6,55221247
edited 12 mins ago
A.K.
6,55221247
6,55221247
asked 2 hours ago
Anonymous I
1115
New contributor
Anonymous I is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Anonymous I
1115
asked 2 hours ago
Anonymous I
1115
1115
New contributor
Anonymous I is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Anonymous I is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Anonymous I is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
@Karl: I meant number of protons in the molecule natriumhydroxide. I changed it.
– Anonymous I
1 hour ago
add a comment |Â
@Karl: I meant number of protons in the molecule natriumhydroxide. I changed it.
– Anonymous I
1 hour ago
@Karl: I meant number of protons in the molecule natriumhydroxide. I changed it.
– Anonymous I
1 hour ago
@Karl: I meant number of protons in the molecule natriumhydroxide. I changed it.
– Anonymous I
1 hour ago
add a comment |Â
1 Answer
1
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up vote
2
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accepted
You need more than a hint. You need a hard shove...
The overall notion here is that water dissociates
$$ceH2O <=> H+ + OH-$$
and that the equilibrium is given by
$$K_w = 1.00times10^-14 = ce[H+][OH-]$$
where $ce[H+]$ is the molar concentration of $ceH+$ ions and $ce[OH-]$ is the molar concentration of $ceOH-$ ions.
NOTE: The pH of a solution is defined as the negative log of the $ceH+$ ion concentration, and pOH of a solution is defined as the negative log of the $ceOH-$ ion concentration. So
$$14 = textpH + pOH$$
Sodium hydroxide is a strong base, so the assumption is that it completely dissociates.
$$ceNaOH(solid) <=> Na+(aq) + OH-(aq)$$
The molecular mass of sodium hydroxide is 40.00 grams/mole.
$$ce[OH-] = dfrac0.60text g40.00text g/moletimes0.80text liter = 0.01875text mole/liter$$
Rearranging the $K_w$ equation
$$ce[H+] = dfracK_wce[OH-] = dfrac1.00times10^-141.875times10^-2 = 5.3times10^-13text mole/liter$$
answered 1 hour ago
MaxW
13.9k12056
Ok, I didn't see that coming. I didn't yet cover that part of chemistry. I'm familiar with the formulas but that's just because I read it from Wikipedia. I'll cover that part now. Thx for the answer though. I think it would be enough if you'd say use pH and the solubility product. My book I used a while ago has a chapter about it.
– Anonymous I
16 mins ago
Hmmm... water doesn't have a "solubility product." There is no precipitate. $K_w$ is known as the (a) ionization constant, (b) dissociation constant, (c) self-ionization constant, or (d) ionic product of water, but never as the solubility product.
– MaxW
3 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You need more than a hint. You need a hard shove...
The overall notion here is that water dissociates
$$ceH2O <=> H+ + OH-$$
and that the equilibrium is given by
$$K_w = 1.00times10^-14 = ce[H+][OH-]$$
where $ce[H+]$ is the molar concentration of $ceH+$ ions and $ce[OH-]$ is the molar concentration of $ceOH-$ ions.
NOTE: The pH of a solution is defined as the negative log of the $ceH+$ ion concentration, and pOH of a solution is defined as the negative log of the $ceOH-$ ion concentration. So
$$14 = textpH + pOH$$
Sodium hydroxide is a strong base, so the assumption is that it completely dissociates.
$$ceNaOH(solid) <=> Na+(aq) + OH-(aq)$$
The molecular mass of sodium hydroxide is 40.00 grams/mole.
$$ce[OH-] = dfrac0.60text g40.00text g/moletimes0.80text liter = 0.01875text mole/liter$$
Rearranging the $K_w$ equation
$$ce[H+] = dfracK_wce[OH-] = dfrac1.00times10^-141.875times10^-2 = 5.3times10^-13text mole/liter$$
answered 1 hour ago
MaxW
13.9k12056
Ok, I didn't see that coming. I didn't yet cover that part of chemistry. I'm familiar with the formulas but that's just because I read it from Wikipedia. I'll cover that part now. Thx for the answer though. I think it would be enough if you'd say use pH and the solubility product. My book I used a while ago has a chapter about it.
– Anonymous I
16 mins ago
Hmmm... water doesn't have a "solubility product." There is no precipitate. $K_w$ is known as the (a) ionization constant, (b) dissociation constant, (c) self-ionization constant, or (d) ionic product of water, but never as the solubility product.
– MaxW
3 mins ago
add a comment |Â
up vote
2
down vote
accepted
You need more than a hint. You need a hard shove...
The overall notion here is that water dissociates
$$ceH2O <=> H+ + OH-$$
and that the equilibrium is given by
$$K_w = 1.00times10^-14 = ce[H+][OH-]$$
where $ce[H+]$ is the molar concentration of $ceH+$ ions and $ce[OH-]$ is the molar concentration of $ceOH-$ ions.
NOTE: The pH of a solution is defined as the negative log of the $ceH+$ ion concentration, and pOH of a solution is defined as the negative log of the $ceOH-$ ion concentration. So
$$14 = textpH + pOH$$
Sodium hydroxide is a strong base, so the assumption is that it completely dissociates.
$$ceNaOH(solid) <=> Na+(aq) + OH-(aq)$$
The molecular mass of sodium hydroxide is 40.00 grams/mole.
$$ce[OH-] = dfrac0.60text g40.00text g/moletimes0.80text liter = 0.01875text mole/liter$$
Rearranging the $K_w$ equation
$$ce[H+] = dfracK_wce[OH-] = dfrac1.00times10^-141.875times10^-2 = 5.3times10^-13text mole/liter$$
answered 1 hour ago
MaxW
13.9k12056
Ok, I didn't see that coming. I didn't yet cover that part of chemistry. I'm familiar with the formulas but that's just because I read it from Wikipedia. I'll cover that part now. Thx for the answer though. I think it would be enough if you'd say use pH and the solubility product. My book I used a while ago has a chapter about it.
– Anonymous I
16 mins ago
Hmmm... water doesn't have a "solubility product." There is no precipitate. $K_w$ is known as the (a) ionization constant, (b) dissociation constant, (c) self-ionization constant, or (d) ionic product of water, but never as the solubility product.
– MaxW
3 mins ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You need more than a hint. You need a hard shove...
The overall notion here is that water dissociates
$$ceH2O <=> H+ + OH-$$
and that the equilibrium is given by
$$K_w = 1.00times10^-14 = ce[H+][OH-]$$
where $ce[H+]$ is the molar concentration of $ceH+$ ions and $ce[OH-]$ is the molar concentration of $ceOH-$ ions.
NOTE: The pH of a solution is defined as the negative log of the $ceH+$ ion concentration, and pOH of a solution is defined as the negative log of the $ceOH-$ ion concentration. So
$$14 = textpH + pOH$$
Sodium hydroxide is a strong base, so the assumption is that it completely dissociates.
$$ceNaOH(solid) <=> Na+(aq) + OH-(aq)$$
The molecular mass of sodium hydroxide is 40.00 grams/mole.
$$ce[OH-] = dfrac0.60text g40.00text g/moletimes0.80text liter = 0.01875text mole/liter$$
Rearranging the $K_w$ equation
$$ce[H+] = dfracK_wce[OH-] = dfrac1.00times10^-141.875times10^-2 = 5.3times10^-13text mole/liter$$
answered 1 hour ago
MaxW
13.9k12056
You need more than a hint. You need a hard shove...
The overall notion here is that water dissociates
$$ceH2O <=> H+ + OH-$$
and that the equilibrium is given by
$$K_w = 1.00times10^-14 = ce[H+][OH-]$$
where $ce[H+]$ is the molar concentration of $ceH+$ ions and $ce[OH-]$ is the molar concentration of $ceOH-$ ions.
NOTE: The pH of a solution is defined as the negative log of the $ceH+$ ion concentration, and pOH of a solution is defined as the negative log of the $ceOH-$ ion concentration. So
$$14 = textpH + pOH$$
Sodium hydroxide is a strong base, so the assumption is that it completely dissociates.
$$ceNaOH(solid) <=> Na+(aq) + OH-(aq)$$
The molecular mass of sodium hydroxide is 40.00 grams/mole.
$$ce[OH-] = dfrac0.60text g40.00text g/moletimes0.80text liter = 0.01875text mole/liter$$
Rearranging the $K_w$ equation
$$ce[H+] = dfracK_wce[OH-] = dfrac1.00times10^-141.875times10^-2 = 5.3times10^-13text mole/liter$$
answered 1 hour ago
MaxW
13.9k12056
edited 1 hour ago
answered 1 hour ago
MaxW
13.9k12056
answered 1 hour ago
MaxW
13.9k12056
answered 1 hour ago
MaxW
13.9k12056
13.9k12056
Ok, I didn't see that coming. I didn't yet cover that part of chemistry. I'm familiar with the formulas but that's just because I read it from Wikipedia. I'll cover that part now. Thx for the answer though. I think it would be enough if you'd say use pH and the solubility product. My book I used a while ago has a chapter about it.
– Anonymous I
16 mins ago
Hmmm... water doesn't have a "solubility product." There is no precipitate. $K_w$ is known as the (a) ionization constant, (b) dissociation constant, (c) self-ionization constant, or (d) ionic product of water, but never as the solubility product.
– MaxW
3 mins ago
add a comment |Â
Ok, I didn't see that coming. I didn't yet cover that part of chemistry. I'm familiar with the formulas but that's just because I read it from Wikipedia. I'll cover that part now. Thx for the answer though. I think it would be enough if you'd say use pH and the solubility product. My book I used a while ago has a chapter about it.
– Anonymous I
16 mins ago
Hmmm... water doesn't have a "solubility product." There is no precipitate. $K_w$ is known as the (a) ionization constant, (b) dissociation constant, (c) self-ionization constant, or (d) ionic product of water, but never as the solubility product.
– MaxW
3 mins ago
Ok, I didn't see that coming. I didn't yet cover that part of chemistry. I'm familiar with the formulas but that's just because I read it from Wikipedia. I'll cover that part now. Thx for the answer though. I think it would be enough if you'd say use pH and the solubility product. My book I used a while ago has a chapter about it.
– Anonymous I
16 mins ago
Ok, I didn't see that coming. I didn't yet cover that part of chemistry. I'm familiar with the formulas but that's just because I read it from Wikipedia. I'll cover that part now. Thx for the answer though. I think it would be enough if you'd say use pH and the solubility product. My book I used a while ago has a chapter about it.
– Anonymous I
16 mins ago
Hmmm... water doesn't have a "solubility product." There is no precipitate. $K_w$ is known as the (a) ionization constant, (b) dissociation constant, (c) self-ionization constant, or (d) ionic product of water, but never as the solubility product.
– MaxW
3 mins ago
Hmmm... water doesn't have a "solubility product." There is no precipitate. $K_w$ is known as the (a) ionization constant, (b) dissociation constant, (c) self-ionization constant, or (d) ionic product of water, but never as the solubility product.
– MaxW
3 mins ago
add a comment |Â
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@Karl: I meant number of protons in the molecule natriumhydroxide. I changed it.
– Anonymous I
1 hour ago