Question

Calculate the ph of the solution resulting by mixing 20.0 ml of 0.15 m hcl with 20.0 ml of 0.10 m koh

Answer 1

Answer:

1.60.

Explanation:

• The no. of millimoles of HCl = MV = (0.15 M)(20.0 mL) = 3.0 mmol.

• The no. of millimoles of KOH = MV = (0.10 M)(20.0 mL) = 2.0 mmol.

Since the no. of millimoles of HCl is larger than that of KOH. The solution is acidic.

∴ M of remaining HCl [H⁺] remaining = (NV)HCl - (NV)KOH/V total = (3.0 mmol) - (2.0 mmol) / (40.0 mL) = 0.025 M.

∵ pH = - log[H⁺]

∴ pH = - log[H⁺] = - log(0.025) = 1.602 ≅ 1.60.

Answer 2

Based on the concentrations and volumes of reactants given, the pH of the resulting solution is 1.60.

What is pH of a solution?

The pH of of a solution is the negative logarithm to base ten of the hydrogen ions concentration of the solution.

• pH = -log[H+]

Also;

• mole of substance = molarity × volume in L

The moles of HCl = 0.15 × 0.02 = 0.003 moles

The moles of KOH = 0.10 × 0.02 = 0.002 moles

The solution will be acidic since the moles of HCl is greater than that of KOH.

Moles of remaining HCl [H⁺] remaining = 0.003 - 0.002 = 0.001 moles

Molarity of remaining HCl [H⁺] remaining = 0.001/0.04 = 0.025 M

Hence, [H⁺] = 0.025 M

pH = - log(0.025)

pH = 1.6

Therefore, the pH of the resulting solution is 1.60.

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