Question

4 stion n an experiment on hydrolysis, the pH values of 0.1M sodium acetate (CHhcoONa) and 0.1 chiorid recorded using pH paper of differentranges. Use these values to determine the correct pH then com Table Abelow DATA AND CALOULATONNS Table A MOH1 Paper Range Wide Range Paper Narrow Range Paper 0.1 MCHvcooNa 4 60 90 95 11 12 Calculations Part B Insert the net ionic equatioms, equilibriumconstant expressions, K. and Ka for the hydrolysable ions, in Table B. Show calculations for K and Ka (see next Pagel. Table B Equilibrium Constant Expression Net Ionic Equation Value 0.1 M NH 12 points) pointsl

Answer 1

Part A:

CH₃COONa is salt of weak acid strong base so its aqueous solution must be basic in nature. pH will be 9.5

NH₄Cl is salt of strong acid weak base so its aqueous solution will be acidic in nature, its pH from above data is 4.5

Now [H⁺] for CH₃COONa =

we have pH= -log₁₀ [H⁺]

So, [H⁺]= antilog[-pH]=antilog(-9.5)= 3.1610^-10

CH_3COONa is salt of weak acid strong base so its aqueous solution must be basic in nature. pH will be 9.5 NH_4Cl is salt of strong acid weak base so its aqueous solution will be acidic in nature, its pH from above data is 4.5 Now [H^+] for CH_3COONa = we have pH= -log_10 [H+] So, [H^+]= antilog[-pH]= antilog(-9.5) = 3.16 times 10^-10 now [H^+][OH^-] = 10^-14 now [OH^-] = 10^-14/3.16 times 10^-10 = 3.16 times 10^-5 For NH_4Cl So, [H^+]= antilog[-pH] = antilog(-4.5) = 3.16 times 10^-5 now [H^+][OH^-] = 10^-14 now [OH^-] = 10^-14/3.16 times 10^-5 = 3.16 times 10^-10

now [H⁺][OH^-]=10^-14 now [OH^-]=10^-14/3.16 $\times$ 10^-10 = 3.16 $\times$ 10^-5

For NH₄Cl

So, [H⁺]= antilog[-pH]=antilog(-4.5)= 3.1610^-5

CH_3COONa is salt of weak acid strong base so its aqueous solution must be basic in nature. pH will be 9.5 NH_4Cl is salt of strong acid weak base so its aqueous solution will be acidic in nature, its pH from above data is 4.5 Now [H^+] for CH_3COONa = we have pH= -log_10 [H+] So, [H^+]= antilog[-pH]= antilog(-9.5) = 3.16 times 10^-10 now [H^+][OH^-] = 10^-14 now [OH^-] = 10^-14/3.16 times 10^-10 = 3.16 times 10^-5 For NH_4Cl So, [H^+]= antilog[-pH] = antilog(-4.5) = 3.16 times 10^-5 now [H^+][OH^-] = 10^-14 now [OH^-] = 10^-14/3.16 times 10^-5 = 3.16 times 10^-10

now [H⁺][OH^-]=10^-14 now [OH^-]=10^-14/3.16 $\times$ 10^-5 = 3.16 $\times$ 10^-10