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# Acid-Base Equilibria Calculated

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1. Calculate [OH-] for 2.170 g of LiOH in 300.0 mL of solution. [OH-] = ? M

2. How many moles of HF (Ka = 6.8 x 10-4) must be present in 0.160 L to form a solution with a pH of 3.30?

3. Calculate the molar concentration of [OH-]ions in a 7.0×10&#8722;2 M solution of ethylamine (C2H5NH2) (Kb= 6.4 x 10-4).

4. Calculate [OH-] and pH for a solution of 7.0 x 10-2 M NaCN? I am not sure if I need Kb or Ka for this equation.

5. What is the pH of a solution that is 3.0×10&#8722;9 M in NaOH? Would I do the log (3.0 x 10-9) and then subtract that number from 14 in order to get the pH?

6. How many milliliters of concentrated hydrochloric acid solution (36.0% HCl by mass, density = 1.18 g/mL) are required to produce 10.0 L of a solution that has a pH of 1.99?

Your help is greatly appreciated!

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#### Solution Preview

1. Calculate [OH-] for 2.170 g of LiOH in 300.0 mL of solution. [OH-] = ? M
Compute for the Molarity of LiOH:
M = moles LiOH / liter solution
M = (mass/molar mass) / liter of solution
M = (2.170 g/23.94 g) / 0.3 L
M = 0.302
Since LiOH is a strong base, it will dissociate 100% in solution. Hence:
LiOH → Li(aq) + OH-(aq)
0.302 M 0.302 M 0.302 M
Therefore, [OH-] = 0.302 M

2. How many moles of HF (Ka = 6.8 x 10-4) must be present in 0.160 L to form a solution with a pH of 3.30?

HF → H+(aq) + F-(aq)
The ratio of H+ and F- per mole of HF that dissociates is 1:1. Hence,
[H+] = [F-]
[H+] = antilog (-pH)
[H+] = antilog (-3.30)
[H+] = 5.01 x 10-4
[F-] = 5.01 x 10-4
Ka = [H+] [F-] / [HF] - [H+]
6.8 x 10-4 = (5.01 x 10-4)( 5.01 x 10-4) / [HF] - 5.01 x 10-4
[HF] = 8.71 x 10-4 M
M = moles/liter solution
8.71 x 10-4 M = moles HF / 0.16 L
Moles HF = ...

#### Solution Summary

The expert calculates acid-base equilibrias. The pH of the solution are determined.

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