Question

How much 6.00 M NaOH must be added to 490.0 mL of a buffer that is 0.0190 M acetic acid and 0.0260 M sodium acetate to raise the pH to 5.75?

Answer 1

the pKa of acetic acid is 4.74.

Now, adding the base to this acidic buffer will cause to increase the concentration of the acetate sodium and decrease the acetic acid; this will cause a pH increase too.

Now in order to do this, we need to work here in moles and use the HH equation so:

pH = pKa - log[A^-]/[HA]

we can work in moles to get a better result (and calculations are easier too)

moles of HA = 0.0190 * 0.490 = 9.31x10^-3 moles
moles of A- = 0.0260 * 0.490 = 0.01274 moles

Now, the base will increase the A- (A- + OH) and decrease the HA (HA - OH), so the value of the volume added will be:

5.75 = 4.74 + log (0.01274 + 6V1 / 0.00931 - 6V1)
1.01 = log(0.01274 + 6V1 / 0.00931 - 6V1)
10.23 = 0.01274 + 6V1 / 0.00931 - 6V1
10.23(0.00931 - 6V1) = 0.01274 + 6V1
0.0952 - 61.38V1 = 0.01274 + 6V1
0.0952 - 0.01274 = 6V1 + 61.38V1
0.08246 = 67.38V1
V1 = 0.001223 L or simply 1.223 mL

Hope this helps