Question

1.Calculate the percent ionization of hydrazoic acid (HN3) in solutions of each of the following concentrations (Ka is given in Appendix D in the textbook). 0.435 M . 0.111 M . 3.79×10−2 M . 2.Consider two solutions, solution A and solution B. [H+] in solution A is 290 times greater than that in solution B. What is the difference in the pH values of the two solutions? Express your answer using two decimal places.

3.What volume of CO2 at 25 ∘C and 1.0 atm is dissolved in a 15.0-L bucket of today's rainwater? Express your answer using two significant figures.

4.What is the conjugate base of HSO3−? Express your answer as a chemical formula.

5.What is the conjugate acid of HPO42− ?

Answer 1

1. From Literature, Ka for HN₃ = 1.9 *10^-5

HN₃ = a M

	HN3	<=>	H+	N3-
I(M)	a		0	0
C(M)	-x		+x	+x
E(M)	a-x		x	x

Ka = [H⁺][N₃^-]/ [HN₃]

1.9*10^-5 = x² / (a-x)

since, Ka is small so x is small can be neglected in comparison to a

So, 1.9*10^-5 = x² / a

(a) a = 0.435 M

1.9*10^-5 = x² / 0.435

x = [H⁺] = 2.87 *10^-3 M

% ionization = ([H⁺] / 0.435)*100% = (2.87 *10^-3 M / 0.435)*100% = 0.66%

(b) a = 0.111 M

1.9*10^-5 = x² / 0.111

x = [H⁺] = 1.45 *10^-3 M

% ionization = ([H⁺] / 0.435)*100% = (1.45 *10^-3 M / 0.111)*100% = 1.31%

(c) a = 3.79*10^-2 M = 0.0379 M

1.9*10^-5 = x² / 0.0379

x = [H⁺] = 8.49 *10^-4 M

% ionization = ([H⁺] / 0.435)*100% = (8.49 *10^-4 M / 0.0379)*100% = 2.24%

2. pH = -log [H⁺]

Let in solution B, [H⁺] = a

In Solution B, [H⁺] = 290 a

so, pH for solution B, pH = -loga

For solution A, pH = -log 290a

Difference in pH = -log a + log 290a + = log290a/290 = 2.46

3.complete information is not given

4. From Brønsted Lowry concept

A conjugate base is formed by the lose proton

Thus, Conjugate base of HSO₃^- is SO₃^2-

5. From Brønsted Lowry concept

A conjugate acid is formed by the gain proton

conjugate acid of HPO₄^2- is H₂PO₄^-