# Acid-Base Equilibria Calculated

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 7.0Ã—10−2 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 3.0Ã—10−9 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 â†’ 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 â†’ 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.