Please answer 1-3 based on the following equilibrium: HX (aq) + H 2 O (l) H 3 O

Question

Please answer 1-3 based on the following equilibrium:

HX (aq) + H2O (l) H3O+ (aq) + X- (aq)

1. In the forward reaction, H2O is behaving as a(n) _________:

a. Bronsted-Lowry acid

b. Bronsted-Lowry base

c. insoluble solute

d. weak acid

e. strong acid

2. If X = F, HX is a ________ and sodium salts of X- are _________:

a. weak acid; neutral

b. weak acid; basic

c. strong acid; neutral

d. strong acid; basic

e. weak acid; acidic

3. Given that for HF (aq) at 25°C, Ka = 3.5 x 10-4, the pKa for HF would be:

a. 3.71

b. 3.22

c. 3.17

d. 3.33

e. 3.46

Explanation / Answer

Solution:- (1) The given equation is....

HX(aq) + H2O(l) <-----> H3O+(aq) + X-(aq)

In this equation, HX is donating a proton, H+ so it is a Bronsted-Lowry acid. This proton is taken by H2O, so water is a Bronsted-Lowry base.

So, the right choice is b.Bronsted-Lowry base.

(2) if X is F then HX will be HF and HF is a weak acid, It's sodim salt would be NaF which is a salt of weak acid(HF) and strong base(NaOH), so the solution would be basic.

The correct choice is b. weak acid; basic.

(3) Pka = - log Ka

Ka is given as 3.5 x 10-4

so, Pka = - log 3.5 x 10-4 = 3.46

The correct choice is e. 3.46

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