Calculate the final pH of a solution made by the addition of 10 mL of a 0.5 M Na

Question

Calculate the final pH of a solution made by the addition of 10 mL of a 0.5 M NaOH solution to 500 mL of a 0.4 M HA originally at pH = 5.0 (pKa = 5.0) Neglect the volume change.

The answer is 5.02

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: the ratio when pka = pH is 1:1. So that means there is .4 M acid and base.

.010L x .5 mol/ L of NaOH: 5 x 10^-3 moles of NaOH

its going into the 500 ml solution so its new molarity is:

5 x 10-3 moles of NaoH / .5 L = .01 M

You add to the base and substract from acid. This is because the acid is reactign with this base. and the base is not reacting with the base, its additive

go to original: .4 M + .01 M = .41 M base

                      .4 M - .01 M = .39 M acid

So pH = 5 + Log (base/acid) = 5 + log (.41/.39) = 5.02

Explanation / Answer

Sol:-

Step 1. Calculation of [ A- ] :-

given [ HA ] = 0.4 M

pH = 5.0 and

pKa = 5.0

Now by using Henderson -Hasselbalch equation , we have

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

5.0 = 5.0 + [ log [ A- ] - log [ HA ] ]

[ log [ A- ] - log [ HA ] ] = 0

log [ A- ] = log [ HA ]  

log [ A- ] = log 0.4

log [ A- ] = - 0.3979

[ A- ] = 10-0.3979

[ A- ] = 0.4 M

Step 2 . Calculation of pH :-

Number of moles of NaOH added = 0.5 M x 0.010 L = 0.005 mol

Number of moles of HA = Number of moles of A- or NaA = 0.4 M x 0.500 L = 0.2 mol

ICF table after the addition of 0.5 M NaOH will be

HA + NaOH <---------> NaA + H2O

I 0.2 mol 0.005 mol 0.2 mol

C - 0.005 - 0.005 + 0.005

F 0.195 mol 0 .00 mol 0.205 mol

again from Henderson-Hasselbalch equation , we have

pH = 5.0 + log 0.205 / 0.195

pH = 5.0 + log 1.05128

pH = 5.0 + 0.02

pH = 5.02

Hence the result  

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