1. How many dissociated H+ ions are there in (3.2x10^-1) L of an aqueous solutio

Question

1. How many dissociated H+ ions are there in (3.2x10^-1) L of an aqueous solution whose pH is 11.55? * Note: your answer is assumed to be reduced to the highest power possible.

2. A volume of 10.0 0 mL of an unknown concentration of acetic acid solution is neutralized by 38.50 mL of 0.236M sodium hydroxide solution. Determine the conventratino of the unknown acetic acid solution.

3. A solution is 0.047 M in nitrous acid, HNO2, has a pH of 4.13. What is the concentration of nitrite ion? (ka of nitrous acid d= 4.5 x10 ^-4)

4. How does a buffer resist change in pH upon addition of a strong acid?

a)The strong acid reacts with the strong base in the buffer to form salt which produces few H+ ions in solution and therefore only a little change in pH.

b) The strong acid reactions with the weak acid in the buffer to form a weak base, which produces few H+ ions in solution and therefore only a little change in pH

c) The strong acid reactions with the weak base in the buffer to form a weak acid, which profuces few H+ ions in solution and therefore only a little change in pH

5. For the acid/base titration of HCl and NaOH, choose a suitable indication from the list below. Beside each indicator, the pH range of colour change at its endpoint is provided.

a) Bromocresol green (3.8-5.4)

b) Brilliant yellow (6.6-7.8)

c) Thymol blue (8.2-9.6)

Explanation / Answer

1. How many dissociated H+ ions are there in (3.2x10^-1) L of an aqueous solution whose pH is 11.55? * Note: your answer is assumed to be reduced to the highest power possible.

by definition:

pH = -log(H+)

so..

[H+] = 10^-pH = 10^-11.55 = 2.8183829*10^-12 M

so V = 0.32 L

then

mol of H+ = MV = (2.8183829*10^-12)(0.32) =9.0188253*10^-13 mol of H+

2. A volume of 10.0 0 mL of an unknown concentration of acetic acid solution is neutralized by 38.50 mL of 0.236M sodium hydroxide solution. Determine the conventratino of the unknown acetic acid solution.

Acetic Acid + NaOH = H2O + NaAcetate

so..

1:1 ratio...

mmol of base = MV = 38.5*0.236 = 9.086 mmol of NAOH

so

9.086 mmol of HAcid ...

M = mmol/mL = 9.086 / 10 = 0.9086 M

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