a) 8.17 b)1.17 c) 5.83 d) 0.82 e) 5.01 2. The pH of a 0.15 M aqueous solution of
ID: 1032036 • Letter: A
Question
a) 8.17
b)1.17
c) 5.83
d) 0.82
e) 5.01
2. The pH of a 0.15 M aqueous solution of NaBrO (the sodium salt of HBrO) is 10.7. What is the Ka for HBrO?
a. 1.3 × 10-12
b. 8.9 × 10-4
c. 6.0 × 10-9
d. 1.6 × 10-6
e. 3.3 × 10-8
3. S is negative for the reaction ________.
a. Sr(NO3)2 (aq) + 2LiOH (aq) Sr(OH)2 (s) + 2LiNO3 (aq)
b. C6H12O6 (s) 6C (s) + 6H2 (g) + 3O2 (g)
c. CH3OH (l) CH3OH (g)
d. LiOH (aq) Li+ (aq) + OH– (aq)
e. 2H2O (g) 2H2 (g) + O2 (g)
4. The addition of HCl and ________ to water produces a buffer solution.
a. HC6H5O
b. HNO3
c. KNO3
d. KOH
e. NH3
a. 2.516
b. 10.158
c. 3.892
d. 4.195
e. 4.502
Explanation / Answer
2. The pH of a 0.15 M aqueous solution of NaBrO (the sodium salt of HBrO) is 10.7. What is the Ka for HBrO?
a. 1.3 × 10-12 b. 8.9 × 10-4 c. 6.0 × 10-9 d. 1.6 × 10-6 e. 3.3 × 10-8
Answer: Since we are asked to find the ka of HOBr, we first write the equation for dissociation of HOBr and find out its formula:
HOBr (aq) ----------------> BrO- (aq) + H+ (aq) -------------1
And ka = [H+][BrO-]/[HOBr] ----------------2
We need concentrations of [H+], [BrO-] & [HOBr], whcih can be obtained based on the given conditions.
The reaction conditions given would be:
NaOBr (aq) + H2O (l) ----------------> HOBr (aq) + OH- (aq) ----------------3
for which the ICE table would be as follows:
NaOBr (aq) + H2O (l) ----------------> HOBr (aq) + OH- (aq)
Initial (I) 0.15 M - 0 0
Change (C) -x - +x +x
Equilibrium (E) (0.15M -x) x x
We are also given the pH of solution, with which we can calculate pOH and hence [OH-] which is same as x in the above equation.
pH = 10.7 ====> -log[H+] = 10.7; [H+] = 10-10.7 =2.00 x 10-11 M
[OH-] = 10-14/ [H+] = 10-14/ 2.00 x 10-11 M ===> [OH-] = 5.01 x 10-4 M
Since x= [OH-] = 5.01 x 10-4 M, [OBr-] at E = [NaOBr] at E = (0.15M - 5.01 x 10-4)====> [OBr-]=0.149 M
similarly according to reaction 3, [HOBr] at E = [OH-] at E = 5.01 x 10-4 M
Subsituting the concentration values for ka in the formula for ka in 2, we get:
ka = [H+][BrO-]/[HOBr] = (2.00 x 10-11 * 0.149)/5.01 x 10-4 = 5.95 x 10-9 which aprroximates to option C.
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