What will happen if a small amount of hydrochloric acid is added to a 0.1 M solu

Question

What will happen if a small amount of hydrochloric acid is added to a 0.1 M solution of HF? a) The percent ionization of HF will increase. b) The percent ionization of HF will decrease. c) The percent ionization of HF will remain unchanged. d) K_a for HF will increase. e) K_a for HF will decrease. Ans: b 15.0 mL of 0.50 M NaOH is added to a 100.-mL sample of 0.442 M NH_3 (k_b for NH_3 1.8 times 10^-5) What is the equilibrium concentration of NH_4^+ ions? a) 1.0 times 10^-2 M b) 6.9 times 10^-6 M c) 3.8 times 10^-1 M d) 1.1 times 10^-4 M e) none of these Ans d Suppose a buffer solution is made from formic acid HCHO_2, and sodium formate, NaCHO_2. What is the net ionic equation for the reaction that occurs when a small amount of sodium hydroxide is added to the buffer? a) NaOH (aq) + H_3O^+ (aq) rightarrow Na^+ (aq) + 2H_2 O(l) b) H_3O^+ (aq) + OH^- (aq) rightarrow 2H_2O(l) c) OH^- (aq) + HCHO_2 (aq) rightarrow CHO_2 (aq) + H_2O(l) d) NaOH (aq) + HCHO_2(aq) rightarrow NaCHO_2 (aq) + H_2O(l) e) Na^+ (aq) + HCHO_2 (aq) rightarrow NaH(aq) + HCO^+_2 (aq) Ans: C

Explanation / Answer

3. The percent ionization of HF will decrease.
For the equilibrium HF(aq) <--> H+(aq) + F-(aq), adding HCl will force the equilibrium to the left, decreasing the %ionization of HF.

6)  Equilibrium: NH3(aq) + NaOH <---> OH-(aq) + NH4+(aq)
Kb = [OH-][NH4+] / [NH3] = 1.8 x10-5

[OH-] from NaOH : 0.015 L (0.50 M) /0.115 L = 0.065 M

[NH3] = 0.1 L (0.442 M) /0.115 = 0.384 M

Ka = [0.065] [NH4+] / [0.384]

1.8 x10-5 x [0.384 ] / [0.065 ] = [NH4+]

[NH4+] = 1.1X10-4 M

12)   In an aqueous solution it is present as a sodium ion (Na+) and a formate ion (CHO2-)

NaOH is added

HCHO2(aq) + NaOH(s) --> NaCHO2(aq) + H2O(l)

ionic equation will be

OH-- (aq) + HCHO2 (aq) ---------> CHO2- (aq) + H2O (l)

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