Respuesta :
Answer: pH=12.69
Explanation:
[tex]{\text{Moles of HF}=Molarity\times {\text{Volume of solution in liters}}[/tex]
[tex]{\text{Moles of HF}=0.20M\times 0.6L=0.12 moles[/tex]
[tex]HF\rightarrow H^++F^-[/tex]
Initial 0.12 0 0
Eqm 0.12-x x x
[tex]K_a=\frac{[H^+][F^-]}{[HF]}[/tex]
[tex]3.5\times 10^{-4}=\frac{x^2}{0.12-x}[/tex]
(neglecting small value of x in comparison to 0.12)
[tex]x=4.2\times 10^{-5}[/tex]
Moles of [tex]H^+=4.2\times 10^{-5}[/tex]
[tex]NaOH\rightarrow Na^++OH^-[/tex]
[tex]{\text{Molesof NaOH}}=Molarity\times {\text{Volume of solution in liters}}[/tex]
[tex]{\text{Moles of NaOH}}=0.10M\times 0.6L=0.06 moles[/tex]
0.06 moles of NaOH will give 0.06 moles of [tex][OH^-][/tex]
Now [tex]4.2\times 10^{-5}[/tex] moles of [tex]OH^-[/tex] will be neutralized by [tex]4.2\times 10^{-5}[/tex] moles of [tex]H^+[/tex] and [tex](0.06-4.2\times 10^{-5})=0.059[/tex] moles of [tex]OH^-[/tex] will be left.
Molarity of [tex]OH^-=\frac{0.059moles}{1.2L}=0.049M[/tex]
[tex]pOH=-\log[OH^-]=-\log[0.049]=1.31[/tex]
pH = 14 - pOH= 14 - 1.31 = 12.69
