Question

What is the pH of a solution of 50.0 mL of 0.112 M NaOH that has been titrated with 36.0 mL of 0.350 M HBr?

Answer 1

no of moles of NaOH = Molarity of NaOH x volume of NaOH in liters

moles of NaOH = 0.112 M x 0.05L = 0.0056 moles

similarly moles of moles of HBr = 0.35 M x 0.036 L = 0.0126

balanced equation is

NaOH + HBr -------> NaBr + H2O

from th ebalanced equation on emole of NaOH consumes one mole of HBr

0.0056 moles of NaOH consumes 0.0056 moles of HBr

remaining moles of HBr = 0.0126 - 0.0056 = 0.007 moles of HBr

molarity of the remaining HBr moles of remaining HBr / total volume

= 0.007 / 0.086 L

= 0.0814 M

since HBr is strong acid concentration of HBr = concentration of H+

pH = -log[H+]

pH = -log[0.0814]

pH = 1.09

=