Let\'s look at a numerical example to demonstrate how adding a strong base to ou

Question

Let's look at a numerical example to demonstrate how adding a strong base to our HCOOH/HCOONa buffer affects the overall phH and how the thermodynamics are operating. We will start with 100.0 mL of a buffer that is 1.04 M HCOOH and 0.960 M HCOONa. The Ks for HCOOH = 6.80 X 10-4 alculating The Initial pH of the Buffer This is just a review of what we learned about the common ion effect. Remember, if there is any weak acid or weak base present, use that equilibrium reaction as your equation. In this case, we have HCOOH, so it's dissociation reaction will be our equation. As we have discussed, the concentration of the [H30+] is not actually equal to zero at the start of the reaction, it is just a convenient (while still accurate) way to solve the problem HCOOH (aq) H20 HCO0 (aq) H30+ (aq) 1.04 M 0.960 M 1.04 x 0.960 x Because the simplifying assumption is valid here we get Ka 6.80 x 104 [(HCOO.(0.960)03 [HCOOH] Therefore: [H30+]s x = 7.37 x 10-4 And the initial pH of the buffer is set at pH = 3.13

Explanation / Answer

The bufferis

HCOOH + HCOO- and pKa of acid = 3.17(given)

a) after 2.2ml of 0.820M KOH is added

HCOOH + OH --------------------> HCOO- + H2O

100x1.04 0 100x0.96 0 initial mmoles

- 2.2x0.82 - - change

98.196 0 101.804 - after

Thus the pH of the buffer is calculated using Hendersen equation

pH = pKa + log [conjugate base]/[acid]

= 3.17 + log 101.804/98.196

= 3.1856

b)If the same 2.2mL of 0.820 M KOH is added to 100mL of water

Now total volume of solution = 100+2.2= 102.2

M1V1 = M2V2

2.2x0.820 = 102.2 x M

Thus molarity of new solution =0.01765 M

Thus pH of solution = 14 -pOH

and pOH = - log 0.01765

=1.7532

and the pH of solution =12.2467

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