Question

A buffer consists of 0.528 M NaH2PO4 and 0.792 M Na2HPO4. Given that the K values for H3PO4 are, Ka1 = 7.2 x 10-3, Ka2 = 6.3 x 10-8, and Ka3 = 4.2 x 10-13, calculate the pH for this buffer.

Answer is 7.38

Not sure how to approach this at all...

Answer 1

@timo86m @oldrin.bataku @nincompoop

If anyone is able to help..I gladly appreciate it! :)

Answer 2

Do you think I just need to use Ka2 since we are only concerned about NaH2PO4 and NaHPO4 ?

Answer 3

I believe so

Answer 4

@oldrin.bataku, That Na2 seems a little tricky now...

Answer 5

don't stress about it; both of those are ionic compounds and dissociate in water

Answer 6

Yeah i see now. its relatively easy ahahah

Answer 7

They key here is

0.528 M NaH2PO4 and 0.792 M Na2HPO4

$$H_2PO_4^+\Longleftrightarrow HPO_4^{2+}+H^+$$Hence we see:$$K_{a}=\frac{[HPO_4^{2+}][H^+]}{[H_2PO_4^+]}\\\log K_a=\log\frac{[HPO_4^{2+}][H^+]}{[H_2PO_4^+]}\\\log K_a=\log\frac{[HPO_4^{2+}]}{[H_2PO_4^+]}+\log[H^+]\\-\log[H^+]=-\log K_a+\log\frac{[HPO_4^{2+}]}{[H_2PO_4^+]}\\pH=pK_a+\log\frac{[HPO_4^{2+}]}{[H_2PO_4^+]}$$