If you mix equal volumes of each of the following pairs of solutions, which will

Question

If you mix equal volumes of each of the following pairs of solutions, which will make a buffer solution? 0.10 M HCl and 0.10 M NaOH 0.50 M HCl and 0.25 M CH_3 CO_2 H 0.50 M NaOH and 0.25 M CH_3 CO_2 H 0.44 M NaOH and 0.88 M CH_5 CO_2 H 0.10 M NaOH and 0.20 M NH_3 The K_a of acetic acid (CH_3 COOH) is 1.8 times 10^-5 If you prepare a solution of 0.25 M sodium acetate (the acid's conjugate base), what is the resulting pH? 4.75 9.26 11.32 9.07 8.15 Which of the following reactions will have an increase in entropy?

Explanation / Answer

44)
acidic buffer is formed by weak acid and salt of weak acid
basic buffer is formed by weak base and salt of weak base

A)
It doesn't form a buffer as it is mixture of strong acid and strong base

B)
It doesn't form a buffer as it is mixture of strong acid and weak acid

C)
CH3COOH will react with NaOH to form CH3COONa but the problem here is all of CH3COOH react and hence finally, it will contain CH3COONa and NaOH. Both are base. this is happening because NaOH is in excess
It doesn't form a buffer

D)
CH3COOH will react with NaOH to form CH3COONa and also NaOH is less
So after raection CH3COONa and CH3COOH will be there
This is buffer

E)
It doesn't form a buffer as it is mixture of strong base and weak base

Answer: D

45)
[CH3COO-] = 0.25 M
Kb of CH3COO- = 10^-14 / Ka
= 10^-14 / (1.8*10^-5)
= 5.56*10^-10

CH3COO- + H2O ----> CH3COOH + OH-
0.25 0 0 (initial)
0.25-x x x (at equilibrium)

Kb = [CH3COOH] [OH-]/[CH3COO-]
5.56*10^-10 = x*x / (0.25-x)

since Kb is small, x will be small and it be ignored as compared to 0.25
above expression thus becomes,

5.56*10^-10 = x*x / (0.25)
x = 1.18*10^-5 M

So,
[OH-] = x = 1.18*10^-5 M

pOH = -log [OH-]
= -log (1.18*10^-5)
= 4.93

pH = 14 - pOH
= 14 - 4.93
= 9.07

Answer: d

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