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### Topic: pH of a buffer.  (Read 2200 times)

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#### Domikss1

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• Mole Snacks: +0/-0 ##### pH of a buffer.
« on: April 12, 2018, 05:27:25 PM »
How many grams of solid phthalic acid (pKa=2.9 pKa2=5.4) are needed to be added to a 100 mL solution of (0.1 M) NaOH to obtain a buffer pH=5.4 ? Answer:1,106g.

I tried using the HH equation to solve this but the answer I ended up with was 0.832g. Could anyone help?

#### Borek ##### Re: pH of a buffer.
« Reply #1 on: April 12, 2018, 05:34:59 PM »
Show how you got 0.832 g.
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#### Domikss1

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• Mole Snacks: +0/-0 ##### Re: pH of a buffer.
« Reply #2 on: April 12, 2018, 07:12:17 PM »
cNaOH=0.1M
V=0.1L
nNaOH= 0.01 mol
MH2A = 166 g/mol

2NaOH + H2 2H2O +Na2A

to neutralize:
2 mol NaOH - 1 mol H2A
0.01 NaOH - x H2A

2x=0.01
x=0.005 mol H2A is needed to neutralize the NaOH producing 0.005 mol Na2A

1) pH = pKa1 + log (nHA-/nH2A)
2) pH = pKa2 + log (nA2-/nHA-)

pKa1 = 2.9
pKa2 = 5.4

1) 5.4 = 2.9 + log (nHA-/nH2A)
2.5 = log (nHA-/nH2A)
102.5 = nHA-/nH2A
102.5·nH2A = nHA-

2) 5.4 = 5.4  + log (nA2-/nHA-)
0 = log (nA2-/nHA-)
100 =nA2-/nHA-
1 = nA2-/nHA-
nHA- = nA2-

nA2- = nNa2A = 0.005 mol

nHA- = nA2- = 0.005

102.5·nH2A = 0.005
nH2A = 0.005 / 102,5 = 0.00001581138 = 1.571138 x 10-5 mol

1.571138 x 10-5 mol + 0.005 mol = 0.00501581138 = 5.01581138 x 10-3 mol

5.01581138 x 10-3 mol x 166 g/mol =0.8326≈ 0.833g.

That's how I calculated it.

#### Borek ##### Re: pH of a buffer.
« Reply #3 on: April 13, 2018, 03:15:53 AM »
To be honest I have no idea what is logic behind your calculations. There is no 0.005 moles of Na2A, as you still have plenty of HA- present in the solution.

You are definitely overcomplicating things. No need to use HH equation twice. Just assume first neutralization goes to completion and the only thing that matters for pH is the A2-/HA- ratio. That's what they did and how they got 1.106 g.
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