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 balanced equation for the reaction is as follows
HBr + KOH --> KBr + H₂O
stoichiometry of HBr to KOH is 1:1
number of moles of HBr reacted - 0.15 mol/L x 0.0500 L = 0.00225 mol 
number of moles of KOH added - 0.25 mol/L x 0.0140 mL = 0.0035 mol
acid to base moles react in the 1:1 ratio
therefore number of KOH moles reacted - 0.00225 mol 
number of KOH moles in excess - 0.0035 - 0.00225 = 0.00125 mol
since KOH is a strong base, it completely ionises
therefore [KOH] = [OH⁻]
[OH⁻] = 0.00125 mol / 0.064 L
[OH⁻] = 0.020 mol/L
we can calculate the pOH 
pOH = -log [OH⁻]
pOH = -log(0.020 M)
pOH = 1.70 

pH + pOH = 14
pH = 14 - 1.70 
pH = 12.3
pH of medium is 12.3

The pH of the solution is calculated as 1.2.

First we obtain the number of moles of each solution as follows;

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)

We can see that the KOH is the limiting reactant and the HBr is present in excess.

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

Molarity of excess acid = 0.004 moles/(50 + 14)× 10^-3 L = 0.0625 M

pH = -log (0.0625 M) = 1.2

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

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