A beaker with 1.80×102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.50 mL of a 0.470 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

The pH will change 0.39

Explanation:

Step 1: Data given

Volume of acetic acid : 180 mL

pH = 5.00

The total molarity of acid and conjugate base in this buffer is 0.100 M.

We add 8.50 mL (= 0.0085 L) of a 0.470 M HCl

pKa of acetic acid = 4.74

Step 2: Calculate initial concentrations

x = concentration acid

y = concentration conjugate base

x + y = 0.100

pH = pKa = log ([y]/[x])

5.00 = 4.74 + log y/x

 0.26 = log y/x

10^0.26 =1.82 = y/x

1.82 x = y

x + 1.82 x = 0.100

2.82 x = 0.100

x = 0.0355 = concentration of acid

y = 0.0645 = concentration of conjugate base

Step 3: Calculate moles

Moles = molarity * volume

moles acid = 0.180 L * 0.0355 M= 0.00639 moles

moles conjugate base = 0.0645 M * 0.180 L=0.01161 moles

moles HCl = 8.50 * 10^-3 L * 0.470 M=0.003995 moles

A- + H+ = HA

moles conjugate base = 0.01161- 0.003995=0.007615  moles

moles acid = 0.00639 + 0.003995=0.010385moles

Step 4: Calculate total volume

total volume = 180 + 8.50 = 188.5 mL = 0.1885 L

Step 5: Calculate concentration

concentration acid = 0.010385 moles/ 0.1885 L =0.0551 M

concentration conjugate base = 0.007615/ 0.1885 =0.0404 M

Step 6: Calculate pH

pH = 4.74 + log 0.0404/0.0551=4.61

Step 7: Calculate pH change

change pH = 5.00 - 4.61=0.39

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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