Question

a) Calculate the pH and pOH for a solution of 0.09952 M NaOH

b) How many acidic protons are there in 0.6137 g of KHP?

c) What volume of 0.1157 M NaOH is required to obtain the endpoint of a reaction with 0.5938 g of KHP?

Answer 1

a. NaOH is strong base

NaOH(aq) -----------> Na^+ (aq) + OH^- (aq)

0.09952M                                     0.09952M

[OH^-]    = [NaOH]

[OH^-]   = 0.09952M

POH   = -log[OH^-]

             = -log0.09952

            = 1.002

PH    = 14-POH

         =14-1.002   = 12.998

b. no of moles of KHP   = W/G.M.Wt

                                     = 0.6137/204.22 = 0.003moles

no of molecules of KHP   = no of moles * 6.023*10^23

                                      = 0.003*6.023*10^23   = 1.807*10^21 molecules

1 molecule of KHP produce 1 acidic proton

1.807*10^21 molecules of KHP produce 1.807*10^21 acidic protons

1.807*10^21 acidic protons >>>>answer

c. KHC4H8O4(aq) + NaOH(aq) --------> KNaC4H8O4(aq) + H2O(l)

   no of moles of KHP   = W/G.M.Wt

                                    = 0.5938/204.22   = 0.0029 moles

KHC4H8O4(aq) + NaOH(aq) --------> KNaC4H8O4(aq) + H2O(l)

1 mole of KHP react with 1 mole of NaOH

0.0029 moles of KHP react with 0.0029 mole of NaOH

no of moles fo NaOH   = molarity * volume in L

                    0.0029       = 0.1157*volume in L

volume in L                 = 0.0029/0.1157   = 0.025L    = 25ml

volume of NaOH = 25ml >>>>answer