1. What is the pH of a solution made by mixing 100.mL each of two solutions of H

Question

1. What is the pH of a solution made by mixing 100.mL each of two solutions of HCl(aq), one with pH = 2.00 and one with pH = 3.00?

2. What volume of 0.150M NaOH is required to reach the endpoint in the titration of 25.00mL of 0.0850M H2C2O4? Note that oxalic acid is diprotic.

3. Indicate whether the each of the following chemical reactions are

A) A Lewis acid/base reaction,

B) An acid/base reaction, but not a Lewis acid/base reaction, or

C) Not an acid/base reaction of any kind.

For each reaction, place one check in the appropriate column. Reaction

A) Lewis a/b

B) a/b

C) not a/b

H+(aq) + OH-(aq) H2O(l)

CO(g) + 2H2(g) CH4(g) + H2O(g)

Ag+(aq) + 2NH3(aq) [Ag(NH3)2]+

HCl(aq) + NaOH H2O(l) + NaCl(aq)

For each reaction, place one check in the appropriate column. Reaction

A) Lewis a/b

B) a/b

C) not a/b

H+(aq) + OH-(aq) H2O(l)

CO(g) + 2H2(g) CH4(g) + H2O(g)

Ag+(aq) + 2NH3(aq) [Ag(NH3)2]+

HCl(aq) + NaOH H2O(l) + NaCl(aq)

Explanation / Answer

1)

We know that pH = -log([H+]

we are mixing two solutions of HCl of pH 2.00 and 3.00 of volume 100 ml each

we need to find the pH of the resulting solution

First we need to find the concentration of [H+] of each solution

pH = 2.00

[H+] = 10-2.00

[H+] = 0.01M

pH = 3.00

[H+] = 10-3.00

[H+] = 0.001M

Concentration of [H+] in final solution is

V1M1 + V2M2 = VM

(100 x 0.01) + (100 x 0.001) = 200 x M

1.1 = 200 xM

M = 1.1 / 200

M =0.0055

[H+] = 0.0055M

pH = -log([H+]

pH = -log(0.0055)

pH = 2.259

2)

2NaOH + H2C2O4 ------> Na2C2O4 + 2H20

The equation show that there are two moles of H+ supplied for each mole of OH-

Let us find the number of moles of H2C2O4

= 25 mL x 0.0850 mmol/mL

= 2.25mmol

2.25mmol has 2 x 2.25 = 4.25 mmol of H+

So to neutralize H2C2O4 we need 4.25 mmol of NaOH

Given is 0.150M of NaOH

Volume of NaOH which has 4.25 mmol of NaOH is (4.25mmol / 0.159 mmol/mL) is 26.73 mL

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