Calculate the pH of a solution obtained by diluting 1 mL of 0.10 M weak monoacidic base to 100 mL at constant temperature if Kb of the base is 1 × 10–5? Solve with calculation. It would be great favour
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Explanation:
V1M1 - V2M2 = V3 M3
50 ×0.2 - 50× 0.1 = 100 × M3
10 - 5 = 100 M3
M3 = 5/100
M3 = 0.05 M
pOH = -log [OH]
pOH = -log 0.05
pOH = 1.3010
pH + pOH = 14
pH = 14 - pOH
pH = 14- 1.3010
pH = 12.699
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