Answer :
The question is incomplete, complete question is :
Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? ([tex]K_a[/tex] for HF is [tex]6.8\times 10^{-4}[/tex].)
[HF] = 0.280 M
Express your answer to two decimal places.
Answer:
The pH of an 0.280 M HF solution is 1.87.
Explanation:3
Initial concentration if HF = c = 0.280 M
Dissociation constant of the HF = [tex]K_a=6.8\times 10^{-4}[/tex]
[tex]HF\rightleftharpoons H^++F^-[/tex]
Initially
c 0 0
At equilibrium :
(c-x) x x
The expression of disassociation constant is given as:
[tex]K_a=\frac{[H^+][F^-]}{[HF]}[/tex]
[tex]K_a=\frac{x\times x}{(c-x)}[/tex]
[tex]6.8\times 10^{-4}=\frac{x^2}{(0.280 M-x)}[/tex]
Solving for x, we get:
x = 0.01346 M
So, the concentration of hydrogen ion at equilibrium is :
[tex][H^+]=x=0.01346 M[/tex]
The pH of the solution is ;
[tex]pH=-\log[H^+]=-\log[0.01346 M]=1.87[/tex]
The pH of an 0.280 M HF solution is 1.87.