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A buffer is created with 0.100 moles of HF and 0.100 moles of NaF in a...

A buffer is created with 0.100 moles of HF and 0.100 moles of NaF in a 1.00 L aqueous solution. This buffer has a pH of 3.15. How many moles of NaOH would you need to add to this buffer in order to obtain a solution with a pH of 3.80. (Ka of HF = 7.1 x 10–4)

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Answer #1

Given that Ka of HF =  7.1 x 10–4

Let us assume , moles of NaOH = x

   HF + NaOH -----------> NaF + H2O

0.1 x x

-------------------------------------------------------------------

0.1-x 0 x

Hence,

pH = -logKa + log [NaF]/[HF]

3.8 = -log (7.1 x 10–4 ) + log [x/0.1-x]

3.8 = 3.15 + log [x/0.1-x]         

x/0.1-x = 1.915

x = 0.1915 - 1.915 x

2.915x = 0.1915

x = 0.066

Therefore,

moles of NaOH required = 0.066 mol

  

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