Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: dennyben on May 06, 2012, 09:59:18 PM
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1) What is the pH of a solution with 10mL of 0.1M NaOH added to 10mL of 0.1M HCl?
2) What is the pH of a solution with 10mL of 0.1M NaOH added to 10mL of 0.1M H2SO4?
3) ... with 10mL of 0.1M NaOH added to 10mL of 0.1M of H3PO4?
I've been looking around for these answers and some say the pHs are 7 and some say otherwise. Some say you need the Ka's and some say you just find -log of concentration?! Please help me out!
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Haha, Ben posted as I was writing this, so there may be some overlap ;D
When an acid is fully reacted with a base, the pH of the solution is determined by the salt created in the reaction.
The salts of strong acids and strong bases are neutral, meaning they don't affect the pH, and so a solution of that salt should have pH=7. For example, HCl + NaOH :rarrow: H2O + Na+ + Cl-, and as NaCl is the salt of a strong acid-strong base reaction, it has no effect on the pH and the pH=7.
The salt of a weak acid-strong base reaction is basic because the acid anion turns into a conjugate base in the salt. For example, H3PO4 + NaOH :rarrow: H2O + Na+ + H2PO4-. The H2PO4- is basic, and so the pH of the solution > 7. You can calculate the pH exactly if you know the concentrations and the Kb of the conjugate base.
The same is true of a strong acid-weak base reaction, except you'll have to use the Ka of the conjugate acid of the weak base. The pH will consequently be less than 7.
Remember that H2SO4 has two acidic hydrogens, so the conjugate base at the first equivalence point is HSO4-. However, as HSO4- can't turn back into sulfuric acid because sulfuric acid ionizes completely, you should still use the Ka of HSO4- to calculate the pH.
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http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt
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Haha, Ben posted as I was writing this, so there may be some overlap ;D
When an acid is fully reacted with a base, the pH of the solution is determined by the salt created in the reaction.
The salts of strong acids and strong bases are neutral, meaning they don't affect the pH, and so a solution of that salt should have pH=7. For example, HCl + NaOH :rarrow: H2O + Na+ + Cl-, and as NaCl is the salt of a strong acid-strong base reaction, it has no effect on the pH and the pH=7.
The salt of a weak acid-strong base reaction is basic because the acid anion turns into a conjugate base in the salt. For example, H3PO4 + NaOH :rarrow: H2O + Na+ + H2PO4-. The H2PO4- is basic, and so the pH of the solution > 7. You can calculate the pH exactly if you know the concentrations and the Kb of the conjugate base.
The same is true of a strong acid-weak base reaction, except you'll have to use the Ka of the conjugate acid of the weak base. The pH will consequently be less than 7.
Remember that H2SO4 has two acidic hydrogens, so the conjugate base at the first equivalence point is HSO4-. However, as HSO4- can't turn back into sulfuric acid because sulfuric acid ionizes completely, you should still use the Ka of HSO4- to calculate the pH.
Thanks for responding! The problems that I have however, are mixtures of acids and bases of the same concentration and amount, meaning the same amount of mols.
Wouldn't this mean that all the OH- ions from the base react and neutralize all the H+ ions from the acid, no matter if they are weak or strong?
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Thanks for responding! The problems that I have however, are mixtures of acids and bases of the same concentration and amount, meaning the same amount of mols.
Wouldn't this mean that all the OH- ions from the base react and neutralize all the H+ ions from the acid, no matter if they are weak or strong?
Yes, but no.
pH is that of a salt solution, so it can be dictated by hydrolysis. Conjugate base of a weak acid is relatively strong, so the solution of sodium acetate will be basic. Conjugate acid of a weak base is a relatively strong acid, so pH of ammonium chloride will be low. And so on.
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I'm still not quite understanding this mathematically.
So let me take a shot at #2 and let me know if it's correct.
2) What is the pH of a solution with 10mL of 0.1M NaOH added to 10mL of 0.1M H2SO4?
(0.1M)(0.01L) = 0.001mol
0.001mol/(0.01L+0.01L) = 0.05M
Ka2 = [SO4][H]/[HSO4]
(1.2x10^-2) = (x^2)/ (0.05)
x = 0.245
-log(0.245) = 1.61 = pH
Is this right?
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Your reasoning look right to me, at first glance. :)
I think you already know this, but just to make it clear, the reason you are starting with HSO4- is because all of the H2SO4 was neutralized by the exact amount of NaOH necessary (0.001 mol NaOH neutralized 0.001 mol H2SO4 to 0.001 mol HSO4-). That's why you are essentially starting out with HSO4- in your Ka equilibrium.
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This is not true
Ka2 = [SO4][H]/[HSO4]
(1.2x10^-2) = (x^2)/ (0.05)
Since HSO4- is an acid with relatively big K you should take this into account:
[SO42-]= 0.05+[H+]
and
[HSO4-]= 0.05-[H+]
My fault. It should be:
[SO42-]= [H+]
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[SO42-]= 0.05+[H+]
No, initially there was no 0.05M of SO42- present.
[HSO4-]= 0.05-[H+]
Yes, without it calculated pH is off by about 0.1.