Question

A 100.0-mL buffer solution is 0.175 M in HClO and 0.150 M in NaClO.Part A: What is the initial pH of this solution?Part B: What is the pH after addition of 150.0 mg of HBr?Part C: What is the pH after addition of 85.0 mg of NaOH?

Answer 1

Answer:

Initial pH of this pH  = 7.453

pH after addition of 150.0 mg of HBr  =  7.35

pH after addition of 85.0 mg of NaOH 0.154  = 7.56

Explanation:

Since Ka value isn't given  

so we use Ka value of HClO (hypo chlorous acid) = 3 x 10⁻⁸

pKa = - logKa = 7.52

Part A

Using Henderson equation

pH = pka + log$\frac{[Conjugate base]}{[Acid]}$

pH = 7.52 + log $\frac{0.15}{0.175}$

pH = 7.453

Part B

pH after addition of 150 mg of HBr

moles of HBr    

             $\frac{Mass}{Molar mass} \\= \frac{150 X10^{-3}g }{80.91} \\= 0.00185 } mole$

Moles of NaOCl in 100 ml buffer solution = $\frac{0.15X100}{1000}$ = 0.015

Moles of HClO in 100 ml buffer solution = $\frac{0.175X100}{1000}$ = 0.0175

Since H⁺ concentration furnished by HBr acid make a common ion effect . So the following reaction carried out

                     

ClO⁻ + H⁺ → HClO

So the remaining concentration of ClO⁻ in solution = 0.015 - 0.000185

                                                                                     = 0.0132            

moles of HClO = 0.0175 + 0.00185

                         = 0.0194

Using Henderson equation pH = Pka + log$\frac{Conjugate base}{Acid}$

                                                    = 7.52 + log$\frac{0.0132}{0.0194}$

                                                     = 7.35

Part C

         pH after addition of 85 mg of HBr

moles of NaOH    

             $\frac{Mass}{Molar mass} \\= \frac{85 X10^{-3}g }{40} \\= 0.00213 } mole$

So the remaining concentration of ClO⁻ in solution = 0.015 + 0.00213

                                                                                     = 0.0171            

Moles of concentration of ClO⁻ = 0.171(M)

moles of HClO = 0.0175 - 0.00213

                         = 0.0154

Moles in 100 ml Buffer = 0.154(M)

Using Henderson equation pH = Pka + log$\frac{Conjugate base}{Acid}$

                                                    = 7.52 + log$\frac{0.171}{0.154}$

                                                     = 7.56