Question

If you start with 500mL solution of 0.02M Na2HPO4 how many mL of 1M NaH2PO4 need to be added to achieve a pH of 7.1? (Please explain yourself, ea chemical reaction would be nice, also the answer should be 50mL)

Answer 1

This is a buffer

if pH is required to be 7.1 then

we need to use Hendesron hasslebach equation

Recall that, in equilibrium

Ka = [H+][A-]/[HA]

Multiply both sides by [HA]

[HA]*Ka = [H+][A-]

take the log(X)

log([HA]*Ka) = log([H+][A-])

log can be separated

log([HA]) + log(Ka) = log([H+]) + log([A-])

note that if we use "pKx" we can get:

pKa = -log(Ka) and pH = -log([H+])

substitute

log([HA]) + log(Ka) = log([H+]) + log([A-])

log([HA]) + -pKa = -pH + log([A-])

manipulate:

pH = pKa + log([A-]) - log([HA])

join logs:

pH = pKa + log([A-]/[HA])

in this case

we need to identify the base and acid

acid --> NaH2PO4 .--> H2PO4-

base = Na2HPO4 --> HPO4-

then

pH = pKa + log([A-]/[HA])

becomes

pH = pKa2 + log([HPO4-2]/[H2PO4-])

pKa2 for this ionization

Ka2 = 6.23*10^-8

pKa = -log( 6.23*10^-8) = 7.21

substitute

7.1= 7.21+ log([HPO4-2]/[H2PO4-])

[HPO4-2]/[H2PO4-] = 10^(7.10-7.21)

[HPO4-2]/[H2PO4-]= 0.776247

initially:

mmol of Na2HPO4 = HPO4-2 = MV = 0.02*500 = 10 mmol

mmol of NaH2PO4 = H2PO4- = 0

in equilbirium

mmol of Na2HPO4 = HPO4-2 = MV = 0.02*500 = 10 mmol

mmol of NaH2PO4  = H2PO4- = MV = 1*V

note that

[HPO4-2]/[H2PO4-]= 0.776247 must be kept

10 / (1*V) = 0.776247

V = 10/0.776247 = 12.882 mL of NaH2PO4 solution

Note that the answer can't be 50 mL

This is too much(proof)

pH= 7.21+ log([HPO4-2]/[H2PO4-])

mmol of HPO4-2 = 0.02*500 = 10

mmol of H2PO4- = 1*50= 50

pH= 7.21+ log(10/50) = 6.51 which is TOO low for the goal of 7.1