Question

A 2.00*10^3 mL buffer solution is 0.500 M in H3PO4 and 0.600 M in NaH2PO4. The Ka of H3PO4 is 7.5*10^-3.

a) What is the initial PH of the buffer solution?

b) What is the pH after the addition of 2.00g of LiOH?

Answer 1

pH of bufer = pKa + log[salt]/[acid]

salt is NaH2PO4 and wea acid is H3PO4

pKa = - log[ka] = - log  7.5*10^-3. = 3 - log7.5 = 2.12

pH = 2.12 + log 0.500/0.600 = 2.12 + 0.079= 2.2

the initial PH of the buffer solution is 2.2

b) concentration of H3PO4 is 0.500 M

M = moles*1000 / volume in mL

0.500 = moles*1000 / 2.00 * 10³

moles = 0.500*2.00 = 1 mole

2.00g of LiOH is how many moles?

moles = wt/ molar mass= 2 g / 24 = 0.083 moles

H3PO4 + LiOH -------> LiH2PO4 + H2O

1 equivalent of LiOH reacts with 1 equiv of H3PO4 and give 1 equiv of salt.

0.083 moles of LiOH reacts with 0.083 moles of H3PO4 and give 0.083 moles of salt.

That means from 1 mole of H3PO4, 0.083 moles of H3PO4 neutalized by LiOH. Then remaining amount of acid is (1 - 0.083) = 0.917 moles.

Then the concentaion of H3PO4 in M = 0.917*1000/ 2.00 * 10³ = 0.4585 M

pH of bufer = pKa + log[salt]/[acid]

pH = 2.12 + log 0.600/0.4585 = 2.12 + 0.1168= 2.24

the pH after the addition of 2.00g of LiOH is 2.24