Question

FAST ANSWER PLEASE FOR BUFFER SOLUTION QUESTION. Would appreciate full working out thanks.

Solution C is made by mixing 55.6 mL of 0.16 M ammonium chloride (NH₄Cl) with 60.3 mL of 0.26 M ammonia solution.

pKa(NH4+) = 9.24

a) Calculate the pH of Solution C.

b) Calculate the pH of the buffer Solution C if 0.14 g of NaOH (molar mass = 40.0 g/mol) is added (assuming no change in volume).

Answer 1

Molarity of NH4Cl = 0.16 M

Volume of NH4Cl = 55.6 mL

Milimoles of NH4Cl = 0.16 * 55.6

= 8.896 mmoles

Molarity of NH3 = 0.26 M

Volume of NH3 = 60.3 mL

Milimoles of NH3 = 0.26 * 60.3

= 15.678 mmoles

pKa(NH4+) = 9.24

Ka = 10^-9.24 = 5.75 x 10^-10

(a). NH4+            H+    +    NH3

        0.16                 0             0
     0.16 - x               x           0.26 + x

Ka = x*(0.26 + x) / (0.16 - x)

5.75 x 10^-10 = x*(0.26 + x) / (0.16 - x)

9.2x10^-11 - 5.75 x 10^-10*x = 0.26*x + x^2

x^2 + 0.26x - 9.2x10^-11 = 0

x = 3.538 x 10^-10 M

[H+] = 3.538 x 10^-10 M

pH = - log [H+]

= - log 3.538 x 10^-10

= 9.45

(b). When .14 g of NaOH (molar mass = 40.0 g/mol) is added

Moles of NaOH added = 0.14 / 40.0

= 3.5 x 10^-3

Total volume of solution = 55.6 + 60.3 = 115.9 mL = 0.1159 L

Moles of H+ ions = 3.538 x 10^-10 * 0.1159

= 4.10 x 10^-11 moles

Moles of OH- in excess = 3.5 x 10^-3 - 4.10 x 10^-11

= 3.499 x 10^-3

[OH-] = 3.499 x 10^-3 / 0.1159

= 0.0301 M

pOH = - log [OH]

= - log 0.0301

= 1.52

pH = 14 - pOH

= 14 - 1.52

= 12.48