Question

How many liters of 0.669 M KOH will be needed to raise the pH of 0.339 L of 4.01 M carbonic acid (H₂CO₃) to a pH of 9.727? Ka=5.6×10–11

Answer 1

first find the moles of each acid and base.

then using the formula calculate the volume of KOH.

moles of H2CO3 = 0.339 L × 4.01 mol/L

= 1.359 mol

moles of KOH = 0.669 × x = 0.669x

now, to draw ICE table.

	H2CO3	H2O	HCO3-	H3O+
initial	1.359	0	0
added	0	0.669x	0
equilibrium	1.359-0.669x	0	0.669x

now the concentration of H2CO3 and HCO3-

[H2CO3]= (1.359-0.669x)/(x+0.339)L

[HCO3-] = 0.669x/(x+0.339)

now to find the volume x using pH formula.

pH = pKa + log([HCO3-]/[H2CO3])

9.727 = - log(5.6 x 10^-11) + log(0.669x/(x+0.339))

10^-0.525 = 0.669x/(x+0.339)

0.299x + 0.101 = 0.669x

x = 0.2729 L