1. Calculate the Ka for an unknown monoprotic acid HX, given that a solution of

ID: 512647 • Letter: 1

Question

1. Calculate the Ka for an unknown monoprotic acid HX, given that a solution of 1.20 M LiX has a pH of 8.90.

4.40

9.60

4.00

4.80

None of these choices are correct.

2. Ka for benzoic acid, C6H5COOH, is 6.30×10-5.

Ka for hypochlorous acid, HClO, is 3.50×10-8.

Ka for hydrocyanic acid, HCN, is 4.00×10-10.

What is the formula for the weakest acid?

3. Calculate the [H+] in a solution that is 0.19 M in NaF and 0.25 M in HF. (Ka = 7.2 × 10–4)

9.5 × 10-4 M

7.2 × 10–4 M

0.20 M

5.5 × 10-4 M

1.3 M

4. 15.0 mL of 0.50 M HCl is added to a 100.-mL sample of 0.480 M HNO2 (Ka for HNO2 = 4.0 × 10-4). What is the equilibrium concentration of NO2- ions?

2.6 × 10-3 M

4.2 × 10-1 M

5.1 × 10-2 M

1.7 × 10-4 M

None of these choices are correct

1. We are given with LiX which will act as base.

So, X- + H2O <---> HX + OH-

we see that concentration of HX will be same as concentration of OH-.

Now, we have pH = 8.90

So, pOH = 14 - 8.90 = 5.10

This give [OH-] = 10^-5.10 =7.94e-6 M

So, [HX] = 7.94e-6M

Kb = [OH-][HX] / [X-] = ( 7.94 x 10^-6)^2 / 1.20 = 5.25e-11

So, Ka = Kw/Kb = 1 e-14 / 5.25e-11 = 1.9e-4.

Now, according to options, looks like the question demands pKa value which will be -log(Ka) = 3.72.

But none of the options match.

So, answer is None of these choices are correct.

Have a nice day!

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