Calculate the pH of a solution with concentrations of 0.50 M formic acid and 0.2
ID: 511633 • Letter: C
Question
Calculate the pH of a solution with concentrations of 0.50 M formic acid and 0.25 M sodium format. K_a = 1.80 times 10^-4 b) To test the buffer capacity of the system in part a (0.50 M formic acid and 0.25 M sodium format), calculate the pH after 0.005 moles Ca(OH)_2 are added to 0.50 L of the buffer; compare this pH value to the pH in part a. c) Compare the pH in part b to the pH of a solution with 0.005 moles Ca(OH)_2 in 0.50 L water (i.e. no buffer) d) Calculate the pH when 0.050 moles HCI are added to 0.50 L of the original buffer in part a (0.50 M formic acid and 0.25 M sodium format).Explanation / Answer
[HA] = 0.50M , [A-] = 0.25 M Ka of acid = 1.8x10-4 and pKa = 3.76
pH of a buffer is given by Hendersen equation as
pH = pKa + log [A-][HA]
= 3.76 + log 0.25/0.50
= 3.46
B) HA + OH- -----------> A-
0.5 L x 0.50 0 0.5x0.25
=0.25 =0.125 initial moles
- 0.005x2 - change
0.24 0 0.135
Now the pH of this buffer
pH = 3.76 + log 0.135/0.24
= 3.51
with the addition of base the ph of buffer slightly increases.
C) [Oh-] in 0.005M ca(OH)2 = 2x0.005 = 0.01 M
thus pOH = -log 0.01 = 2
the pHof this solution = 14 -pOH
= 14- 2= 12
D)
HA -----------> A- +HCl
0.5 L x 0.50 0.5x0.25
=0.25 =0.125 0 initial moles
- - 0.05 change
0.30 0.075 after rxn
Thus the pH of the new buffer is
pH = 3.76 + log 0.075/0.30
=3.1579
Thus on addition of HCl , the pH decreased.
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