Question

The temperature for each solution is carried out at approximately 297 K where Kw=1.00×10−14.

Part A

0.30 g of hydrogen chloride (HCl) is dissolved in water to make 4.0 L of solution. What is the pH of the resulting hydrochloric acid solution?

Express the pH numerically to two decimal places.

Part B

0.80 g of sodium hydroxide (NaOH) pellets are dissolved in water to make 3.0 L of solution. What is the pH of this solution?

Express the pH numerically to two decimal places.

Ammonia, NH3, is a weak base with a Kb value of 1.8×10−5.

Part C

What is the pH of a 0.155 M ammonia solution?

Express your answer numerically to two decimal places.

Part D

What is the percent ionization of ammonia at this concentration?

Express your answer with the appropriate units.

Answer 1

A)

[HCl] = mol/V = (mass/MW) / V = (0.3/36.5) / 4 = 0.00205

[H+] = [HCl] = 0.00205

pH = -log(0.00205 =2.6882

pH= 2.69

b)

mol of NaOH = mass/M W= 0.8/40 = 0.02

[NaOH] = mol/V = 0.02/3 = 0.006666

[OH-] = 0.006666

pOH = -log(0.006666) = 2.176

p´H = 14-2.176

pH = 11.824

c)

This is a base in water so, let the base be CH3NH2 = "B" and CH3NH3+ = HB+ the protonated base "HB+"

there are free OH- ions so, expect a basic pH

B + H2O <-> HB+ + OH-

The equilibrium Kb:

Kb = [HB+][OH-]/[B]

initially:

[HB+] = 0

[OH-] = 0

[B] = M

the change

[HB+] = x

[OH-] = x

[B] = - x

in equilibrium

[HB+] = 0 + x

[OH-] = 0 + x

[B] = M - x

Now substitute in Kb

Kb = [HB+][OH-]/[B]

Kb = x*x/(M-x)

x^2 + Kbx - M*Kb = 0

x^2 + (1.8*10^-5)x - (0.155)(1.8*10^-5) = 0

solve for x

x = 0.00166

substitute:

[OH-] = 0.00166

pH = 14 + pOH = 14 + log(0.00166) = 11.22

d)

%ionizatoin = [OH-]/[NH3] initial * 100% =(0.00166)/(0.155) * 100% = 1.070%