Question

Calculate the pH of a solution after 10.0mL of .1M NaOH is added to 40mL of .250M HBr.

Answer 1

let us first write the data provided

Molarity of NaOH =0.1 M               Volume of NaOH =10 ml =10*10^-3 Litre

Molarity of HBr =0.25 M               Volume of HBr =40 ml = 40 *10^-3 Litre

Number of moles of NaOH= Molarity * Volume =0.1*10*10^-3 =0.001 moles

Number of moles of HBr = Molarity *Volume =0.25*40*10^-3=0.01 moles

The reaction can be written as

                                      NaOH + HBr ------>      NaBr   +    H20

                  at time t=0      0.001      0.01            -           -

                 at time t=t        -           0.01-0.001      0.001 -

                                                       =0.009

Hence excess of HBr present =0.009 moles . This will contribute to the pH of the solution.

Volume of final Solution = 10+40 =50 ml =50*10^-3 Litre

Hence concentration of excess HBr present =0.009/(50*10^-3) = 0.18 M

Therefore pH of solution is, pH=-Log([HBr]) = - Log(0.18) =0.7447 ,

Therefore the pH of the solution is 0.7447.