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?

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

pH = 7.52 + log

pH = 7.453

Part B

pH after addition of 150 mg of HBr

moles of HBr    

Moles of NaOCl in 100 ml buffer solution = = 0.015

Moles of HClO in 100 ml buffer solution = = 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

                                                    = 7.52 + log

                                                     = 7.35

Part C

         pH after addition of 85 mg of HBr

moles of NaOH    

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

                                                    = 7.52 + log

                                                     = 7.56

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b