Question

9. Write the B/L acid-base reaction between ammonium ion and water. Is ammonium ion acting as an acid or a base in this reaction? Does the equilibrium lie to the product or the reactant side? 10. Write the B/L acid base reaction between acetate ion and bisulfate ion. Identify the two acid and base conjugate pairs above reaction. Does the equilibrium lie to the product or the reactant side? 11. Calculate the pH if: (report values to 2 decimal places) [H"]=4.2 x 10-M [H*] = 5.7 x 10M [H*] =3.6 x 10-PM [H] = 1.2 x 10-12 M 12. Calculate the pOH if: (report values to 2 decimal places) [H*] = 4.2 x 10-M [H*] = 5.7 x 10M [H*] = 3.6 x 10°M [H] = 1.2 x 10-12 M 13. Calculate the [H3O+] if: (report values to 2 sig. figs.) [OH-] = 2.4 x 10M [OH-] = 6.8 x 10-8 M [OH-] = 8.6 x 10-'M [OH-] = 3.1 x 10M 14. Calculate the [pH] if: (report values to 2 sig. figs.) [OH-] = 4.5 x 10M [OH-] = 8.2 x 10-8 M [OH-] = 3.7 x 10'M [OH-] = 6.9 x 10-12 M 15. What is the pH of a buffer that contains 0.120M HF and 0.100M NaF? The Ka of HF is 3.5x104.

Answer 1

Question 11)

Since pH of any solution is calculated by fomula

pH = -log[H₃O⁺] or pH = -log[H⁺]

a) For 4.2*10^-4 M

pH = -log(4.2*10^-4)

= 3.3767 = 3.38

Hence pH of solution will be 3.38

b) pH = -log(5.7*10^-6)

= 5.2441 = 5.24

Hence pH of solution will be 5.24

c) pH = -log(3.6*10^-9)

= 8.4436 = 8.44

Hence pH of solution will be 8.44

d) pH = -log(1.2*10^-12)

= 11.9208 = 11.92

Hence pH of solution will be 11.92

In multiple questions we solve only one question at a time. Thanks