Question

Would pls show all steps

If 200 ml of 0.16 M phosphate buffer at pH 7.2 has a new pH of 6.9 after some volume of 0.12 M hydrochloric was added. what volume of 0.12 M of hydrochloric acid was added.

Answer 1

total no of mol of phosphatebuffer = 0.16*0.2 = 0.032 mol

pH of phosphate buffer = pka2 + log[HPO4^-2/H2PO4^-]

pka2 of H3PO4 = 7.2

[HPO4^-2] = x

[H2PO4^-] = 0.032-x

7.2 = 7.2 + log(x/(0.032-x))

x = 0.016

[HPO4^-2] = x = 0.016

[H2PO4^-] = 0.032-x = 0.032-0.016 = 0.016

after addition of HCl,

pH = pka2 + log[HPO4^-2 - acid]/[H2PO4^- + acid]

6.9 = 7.2 + log((0.016-x)/(0.016+x))

x = 0.0053

no of mole of HCl added = x = 0.0053

volume of HCl added = 0.0053/0.12 = 0.0442 L

                    = 44.2 ml