Question

Calculate the pH of a buffer solution that contains 0.71 M NaH₂PO₄ and 0.12M Na₂HPO_4.

_{I got the answer to the first part which is 6.44, I need help with the second part}

Calculate the change in pH if 0.070 g of solid NaOH is added to 250 mL of the solution in the problem above
.

Answer 1

Calculate the pH of a buffer solution that contains 0.71 M NaH₂PO₄ and 0.12M Na₂HPO_4.

_{I got the answer to the first part which is 6.44, I need help with the second part}

Calculate the change in pH if 0.070 g of solid NaOH is added to 250 mL of the solution in the problem above

NaH₂PO₄ -------- > Na⁺+ H₂PO^-₄

Na₂HPO₄ -------- > 2Na⁺+ HPO^2-₄

For NaH₂PO₄ / Na₂HPO₄ buffer we have the equilibrium as,

H₂PO^-₄ -------- > H⁺ + HPO^2-₄

We initially have,

[] = [] =

[NaH₂PO₄] =[H₂PO^-₄] = [acid] = 0.71 M

[Na₂HPO₄]=[HPO^2-₄] =[salt] = 0.12 M

And pH = 6.44

Let us calculate, pKa for (H₂PO^-₄) = ?

Henderson’s equation,

pH = pKa + log{[salt]/[acid]}

pH = pKa + log{[HPO^2-₄]/ [H₂PO^-₄]}

6.44 = pKa + log(0.12/0.71)

6.44 = pKa + log(0.169)

6.44 = pKa +(- 0.77)

pKa = 6.44 + 0.77

pKa = 7.21.

pKa for (H₂PO^-₄) =7.21. (Note this value we can also take from literature directly)

Now let us calculate moles/L of NaOH added externally to this buffer,

We know that,

Molar mass of NaOH = 40 g/mol

Means 40 g NaOH in net 1000 mL solution makes 1M/L NaOH solution,

i.e. 40 g NaOH = 1 M/L in 1000 mL

40 g NaOH = (¼) M/L in 250 mL……..(Divided by 4 on both side for 250 mL volume)

Then, 0.07 g NaOH = say ‘ A’ M/L

A = [(1/4 ) x 0.07] / 40

A = 4.375 x 10^-4 M/L

i.e. we are adding 0.4375 M/L of NaOH externally

This added concentration of NaOH will change concentartions as follows,

New [HPO^2-₄] =[salt] = Initial conc x volume – Conc. Of NaOH added

                                        = 0.12 M x(1/4 L)- 4.375 x 10^-4 M/L

                                        = 0.030 M/L

[H₂PO^-₄] = [acid] = 0.71 M x(1/4 L)- 4.375 x 10^-4 M/L =0.177 M/L

Using Hendeson equation let us calculate new pH

pH = pKa + log{[salt]/[acid]}

pH = pKa + log{[HPO^2-₄]/ [H₂PO^-₄]}

pH = 7.21 + log(0.03/0.177)

pH = 7.21 + log(0.1694)

pH = 7.21 +(-0.7710)

pH = 7.21 - 0.7710

pKa = 6.439.

I.e. There will be almost no change in original pH.

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