Chemistry 2 The pOH of a solution at 25 degree C in which [H degree] = 2.45 time

Question

Chemistry 2 The pOH of a solution at 25 degree C in which [H degree] = 2.45 times 10^8 M is: Impossible to determine 7.61 6.39 7.16 6.84 A buffer solution contains 0.400 mol HNO_2 and 0.400 NaNO_2 in a 1.00 L solution. (K_a for HNO_2 is 4.5 times 10^-4). A small amount of NaOH (0.050 moles) is added to the buffer solution. What is the pH after addition of NaOH? 3.46 3.35 3.24 3.50 4.50 Find the percent dissociation (aka percent ionization) of 0.455 M phenol (C_6 H_5 OH; K-a = 1.3 times 10^-10) 5.92 times 10^-11 7.69 times 10^-6 1.69 times 10^-5 1.69 times 10^-3 7.69 times 10^-4

Explanation / Answer

Q19

first, find pH:

pH = -log(H+)

pH = -log(2.45*10^-8)

pH = 7.610

so..

at 25°C, we know that

14 = pH + pOH

so

pOH = 14-pH

pOH = 14-7.610

pOH = 6.39

which is basic, choose C

Q.20

first, identify the amount of acid and base present

mmol of acid = 0.4

mmol of conjugate = 0.4

after adding 0.05 mmol of strong base, expect

HNO2 + OH- = H2O + NO2 fromation

so: acid decreases, conjugate is formed

mmol of acid = 0.4 - 0.05 = 0.35

mmol of conjugate = 0.4 +0.05 = 0.45

substitute in pH equation of buffer

pH = pKa + log(NO2/HNO2)

pKa = -log(Ka) = -log(4.5*10^-4) =3.346

pH = pKa + log(NO2/HNO2)

pH = 3.346 + log(0.45/0.35)

pH = 3.45514

choose a

Q21

% dissociation for acid = [H+]/M * 100%

so...

M = 0.455

and

[H+] = from dissociation constant.

Ka = [H+][C6H5O-]/[C6H5OH]

1.3*10^-10 = x*x/(0.455-x)

x = 7.69*10^-6

so

[H+] = 7.69*10^-6

%dissociation = (7.69*10^-6) / (0.455) * 100 = 0.00169 %

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