use:
pKa = -log Ka
4.82 = -log Ka
Ka = 1.514*10^-5
Given:
M(HC3H7CO2) = 1.1 M
V(HC3H7CO2) = 138.2 mL
M(KOH) = 0.6 M
V(KOH) = 294.3 mL
mol(HC3H7CO2) = M(HC3H7CO2) * V(HC3H7CO2)
mol(HC3H7CO2) = 1.1 M * 138.2 mL = 152.02 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.6 M * 294.3 mL = 176.58 mmol
We have:
mol(HC3H7CO2) = 152.02 mmol
mol(KOH) = 176.58 mmol
152.02 mmol of both will react
excess KOH remaining = 24.56 mmol
Volume of Solution = 138.2 + 294.3 = 432.5 mL
[OH-] = 24.56 mmol/432.5 mL = 0.0568 M
use:
pOH = -log [OH-]
= -log (5.679*10^-2)
= 1.2458
use:
PH = 14 - pOH
= 14 - 1.2458
= 12.7542
Answer: 12.75
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