Question

1. A buffer solution contains 0.257 M NaHCO₃ and 0.498 M K₂CO₃.
If 0.0231 moles of hydrochloric acid are added to 125 mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume change does not change upon adding hydrochloric acid)
pH =

1A. A buffer solution contains 0.303 M NaH₂PO₄ and 0.307 M Na₂HPO₄.
If 0.0440 moles of potassium hydroxide are added to 250. mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume does not change upon adding potassium hydroxide)

pH =

Answer 1

NaHCO3 = 0.257M

K2CO3 = 0.498M

Volume of the solution = 125 ml = 0.125 L

number of moles of NaHCO3 = 0.257M x 0.125L = 0.032125 moles

number of moles of K2CO3 = 0.498Mx0.125L =0.06225 moles

number of moles of Hydrochloric acid = HCl = 0.0231 moles

after addition of HCl

number of moles of NaHCO3 = 0.032125 + 0.0231 = 0.055225 moles

number of moles of K2CO3 = 0.06225 - 0.0231 = 0.03915 moles

Ka of HCO3- = 4.8x10^-11

-log(Ka) = -log( 4.8x10^-11)

PKa = 10.32

PH = Pka + log[salt/acid]

PH = 10.32 + log( 0.03915/0.055225)

PH =10.17.

2)

NaH2PO4 = 0.303M

Na2HPO4= 0.307M

volume of the solution = 250 ml = 0.250L

number of moles fo NaH2PO4 = 0.303Mx0.250L = 0.07575 moles

number of moels of Na2HPO4 = 0.307M x0.250L= 0.07675 moles

Ka of H2PO4- = 6.23x10^-8

log(ka) = -log( 6.23x10^-8)

Pka= 7.2

number of moles ofKOH = 0.0440 moles

after addition of KOH

number of moles of NaH2PO4 = 0.07575 - 0.0440=0.03175 moles

number o fmoles of Na2HPO4 = 0.07675 + 0.0440 = 0.12075 moles

PH = 7.2 + log( 0.12075/0.03175)

PH = 7.78