Calculate the [OH^-] concentration in an aqueous solution that is 3.50 times 10^
ID: 525354 • Letter: C
Question
Calculate the [OH^-] concentration in an aqueous solution that is 3.50 times 10^-5 M in H^+ A) 2.86 times 10^-2 M B) 2.86 times 10^-9 M C) 2.86 times 10^-10 M D) 3.50 times 10^-10 M Calculate the pH for an aqueous solution is 2.15 times 10^-4 M in OH^- A) 4.65 times 10^-11 B) 2.15 times 10^-4 C) 10.33 D)3.67 Which of the following acids is the WEAKEST? The acid is followed by its K_a value. A) HCIO_2, 1.1 times 10^-2 M B) HCN, 4.9 times 10^-10 M C) HNO_2, 4.6 times 10^-4 M D) C_6H_5COOH, 6.5 times 10^-5 M Determine the pH of a 0.290 M Mg(OH)_2 solution at 25 degree C. A) 13.46 B) 13.76 C) 0.537 D) 12.92 E) 0.237 You wish to prepare an HC_2H_3O_2 buffer with a pH of 4.14. If the pK_a of is 4.74, what ratio of C_2H_3O_2^-/HC_2H_3O_2 must you use? A) 0.25 B) 0.10 C) 3.98 D) 4.0 E) 0.60 How many milliliters of 0.120 M NaOH are required to titrate 50.0 mL of 0.0998 M hypochlorous acid to the equivalence point? The K_a of hypochlorous acid is 3.0 times 10^-8 A) 41.6 B)7.60 C)60.1 D) 50.0 E) 7.35 Give the expression for the solubility product constant for BaF_2 K_sp = ___ A) [Ba^2+] [F^-] B) [Ba^2+]^2 [F^-] C) [Ba^2+] [F^-]^2 D) [Ba^2+][F^-]^2/[BaF_2]Explanation / Answer
8. C. [H+] = 3.50 x 10-5 M [OH-] = 10-14/[H+] = 10-14/ 3.50 x 10-5 = 2.86 x 10-10 M
9. C.
[OH-] = 2.15 x 10-4 M ; [H+] = 10-14/[OH-] = 10-14/ 2.15 x 10-4 = 4.65 x 10-11 M
PH = - log [H+] = - log (4.65 x 10-11 ) = 10.33
10. B. HCN Lower the value of Ka, lower is the concentraion of H+ ion in solution and weaker is the acid
11. B
Mg(OH)2 <--------> Mg2+ + 2 OH-
0.290 M will produce 2 x 0.290 M = 0.58 M OH- ion
[H+] = 10-14/[OH-] = 10-14/ 0.58 = 1.72 x 10-14 M
PH = - log [H+] = -log (1.72 x 10-14) = 13.76
12. A. 0.25
PH = Pka + log {[C2H3O2-]/[HC2H3O2]
4.14 = 4.74 + log {[C2H3O2-]/[HC2H3O2]
log {[C2H3O2-]/[HC2H3O2] = - 0.60
[C2H3O2-]/[HC2H3O2] = antilog (- 0.60) = 0.25
13. A. 41.6 mL ( V1M1 = V2M2 where V1 and V2 are volume of acid and base respectively ; M1 and M2 are the molarity of acid and base respectivly)
volume of NaOH = (50 x 0.0998)/0.120 = 41.6 mL
14. C.
BaF2 <------> Ba2+ + 2F-
Ksp = [Ba2+][F-]2