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Question 22. A buffer is created using acetic acid (HC_2H_3O_2) and sodium aceta

ID: 944760 • Letter: Q

Question

Question 22. A buffer is created using acetic acid (HC_2H_3O_2) and sodium acetate (NaC_2 H_3 O_2).If small amount of NaOH is added, which species will react with the hydroxide (OH^-) to keep pH from changing?H_2O Na^+ HC_2 H_3 O_2 H_3O^4 C_2H_3O_2 Consider a 0.1 M solution of HF. What will happen if some sodium fluoride salt (NaF) is added to the solution? The percent ionization of HF will increase. The k _a value for HF will increase. The percent ionization of HF will remain unchanged. The K _a value for HF will decrease. The percent ionization of HF will decrease. Which of these options will form the most effective buffer system with a pH value near 6.30? H_3PO_4 and H_2PO_4 (K _al for H_3PO_4 is 6.9 times 10^-3) H_2CO_3 and HCO^-_3 (K _al for H_2CO_3 is 4.3 times 10^-7) H_2S and HS^- (K _al for H_2S is 8.9 times 10^-8) HCO^-_3 and C0^2- _3 (K_a2 for H_2CO_3 is 4.8 times 10^-11) HC_2O^- _4-and ( K_a2 for H_2C_2O_4 is 5.1 times 10^-5)

Explanation / Answer

22)

[HF] = 0.1

when we add NaF...

we will have a buffer since there will be

HF <--> H+ and F- (in equilibrium)

NaF --> Na+ and F- (the common ion effect)

therefore

[F-] ion in solution increases, then the products increase.... accordint to lechatelier, the equilibrium will be shifted towaard reactants

that is

HF

then

HF increases

% ioniziaton = H+ / HF * 100

if H decrases and HF increases, then % ionization decreases

choose E

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