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The pH of a 1. 0 l buffer prepared by adding 0. 100 moles of NaOH to 0. 250 moles of HF. ka for HF = 3. 5 × 10−4 is 3.86.

What is Henderson hasselbalch equation?

The Henderson hasselbalch equation can be expressed as

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

Firstly we will calculate the value of pKa

pKa = -logKa

Given,

Ka = 3. 5 × 10−4)

pKa = -log(3. 5 × 10−4)

pKa = 3.46

Now, we will calculate the value of log [base]/ [acid]

Given,

[base] = 0.10m

[acid] = 0.25m

log [base]/ [acid] = log(0.10/0.25)

= 0.4

Putting the values in handerson hasselbalch equation,

pH = 3.46+0.4

= 3.86

Thus, we find that the value of pH of a 1. 0 l buffer prepared by adding 0. 100 moles of NaOH to 0. 250 moles of HF is 3.86.

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