Calculate the pH of a solution obtained by mixing 25.00 mL of 0.13 M NH3 with 25.00 mL of 5.0×10-2 M HCl. Kb for NH3 is 1.8×10-5.
NH3 concentration = 0.13, i.e. 0.13 moles of NH3 in 1000 mL of soution, therefore no. of moles of NH3 in 25 mL = 0.13/1000 * 25 = 0.00325.
similarly moles of HCl = 5 x 10-2 x 0.025 = 0.000125.
Since NH3 is less, HCl will be completely neutralized and some amount of unreacted NH3 will be left.
so the resulting solution of 50 mL has 0.000125 moles of NH4Cl and (0.00325 - 0.000125) 0.003125 moles of NH3.
therefore concentration of NH4Cl = 0.000125/50 * 1000 = 0.0025M
concentration of NH3 = 0.003125/50 * 1000 = 0.0625M
pOH = pKb + log [salt]/[base]
pKb = - log Kb = 4.744
pOH = 4.744 + log 0.0025/0.0625 = 3.34
pH = 14 - pOH = 14 - 3.34 = 10.66
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