Question

Determine the pH of the following solution:

A 1.0 L of an aqueous solution containing 3.0 grams of HBr and 7.0 mL of 1.0 M NaOH.

Answer 1

no of moles of HBr = weight/ molar mass of the HBr = 3.0 g / 79.93 g/mol = 0.03753 moles

no of moles of NaOH = molarity of the NaOH x volume of NaOH in liters = 1.0M x 0.007L = 0.007 moles

total volume of the solution = 1.0L

now write the balanced equation between them

HBr + NaOH ----> NaBr + H2O

from this equation it is clear that one mole of HBr will consume one mol eof NaOH

here NaOH is limiting agent so

0.007 moles of NaOH will consume the 0.007 moles of HBr

no o fmoles of HBr remaining = 0.03753 - 0.007 = 0.03053 moles of HBr remaining

concentration of HBr remaining = moles of HBr remaining / volume of the solution in liters

= 0.03053 moles / 1.0 L

= 0.03053 mol/L

since HBr is a strong acid we can take the concentration of HBr = concentration of H+ = 0.03053

pH = -log{H+]

pH = -log[0.03053]

pH = 1.5152