A50.0-ml volume of 0.15 m hbr is titrated with 0.25 m koh. calculate the ph after the addition of 14.0 ml of koh. express your answer numerically.

The pH of the solution is calculated as 1.2.

Number of moles of HBr = 50/1000 × 0.15 M = 0.0075 moles

Number of moles of KOH = 14/1000 × 0.25 M = 0.0035 moles

Equation of the reaction; HBr(aq) + KOH(aq) > KBr(aq) + H2O(l)

Number of moles of excess HBr =  0.0075 moles - 0.0035 moles = 0.004 moles

pH = -log (0.0625 M) = 1.2

Hence, the pH of the solution is calculated as 1.2.

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explanation:

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