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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: Domikss1 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?

Show how you got 0.832 g.

c_{NaOH}=0.1M
V=0.1L
n_{NaOH}= 0.01 mol
M_{H2A} = 166 g/mol
2NaOH + H_{2}A :rarrow: 2H_{2}O +Na_{2}A
to neutralize:
2 mol NaOH  1 mol H_{2}A
0.01 NaOH  x H_{2}A
2x=0.01
x=0.005 mol H_{2}A is needed to neutralize the NaOH producing 0.005 mol Na_{2}A
1) pH = pKa_{1} + log (n_{HA}^{}/n_{H2A})
2) pH = pKa_{2} + log (n_{A}^{2}/n_{HA}^{})
pKa_{1} = 2.9
pKa_{2} = 5.4
1) 5.4 = 2.9 + log (n_{HA}^{}/n_{H2A})
2.5 = log (n_{HA}^{}/n_{H2A})
10^{2.5} = n_{HA}^{}/n_{H2A}
10^{2.5}·n_{H2A} = n_{HA}^{}
2) 5.4 = 5.4 + log (n_{A}^{2}/n_{HA}^{})
0 = log (n_{A}^{2}/n_{HA}^{})
10^{0} =n_{A}^{2}/n_{HA}^{}
1 = n_{A}^{2}/n_{HA}^{}
n_{HA}^{} = n_{A}^{2}
n_{A}^{2} = n_{Na2A} = 0.005 mol
n_{HA}^{} = n_{A}^{2} = 0.005
10^{2.5}·n_{H2A} = 0.005
n_{H2A} = 0.005 / 10^{2,5} = 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.

To be honest I have no idea what is logic behind your calculations. There is no 0.005 moles of Na_{2}A, 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 A^{2}/HA^{} ratio. That's what they did and how they got 1.106 g.