Answer:
12.6.
Explanation:
no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.
no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.
So,
[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.
∵ pOH = -log[OH⁻]
∴ pOH = -log(0.395 M) = 1.4.
∵ pH + pOH = 14.
∴ pH = 14 - pOH = 14 - 1.4 = 12.6.
12.6 is the pH of a solution prepared by mixing 45.0 ml of 0.183 m KOH and 35.0 ml of 0.145 m HCl.
pH of the solution can be calculated as pH = 14 - pOH.
First of all we have to calculate the concentration of OH⁻ ions as follow:
Total volume of the solution = 45ml + 35ml = 80ml
Concentration of OH⁻ in terms of mole is calculated as:
[OH⁻]=(no. of millimoles of KOH - no. of millimoles of HCl)/Total volume
Given that, concentration of KOH = 0.183m
Volume of KOH = 45ml
Concentration of HCl = 0.145m
Volume of HCl = 35ml
Millimoles can be calculated as:
millimoles = concentration × volume
No. of millimoles of KOH = 0.183 M × 45.0 mL = 8.235 mmol
No. of millimoles of HCl = 0.145 M ×35.0 mL = 5.075 mmol
Now, [OH⁻] = 8.235 mmol - 5.075 mmol / 80.0 mL = 0.395 M
pOH = -log[OH⁻]
pOH = -log(0.395)
pOH = 1.4
Therefore, pH = 14 - 1.4 = 12.6
Hence, 12.6 is the pH of the solution.
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