A 1.00L buffer solution is 0.250M HF and 0.250M LiF. Calculate the pH of this so

Question

A 1.00L buffer solution is 0.250M HF and 0.250M LiF. Calculate the pH of this solution after addition of 0.150 moles of solid LiOH. Ka HF=3.5x10^-4.

A solution contains 0.021M Cl- and 0.017M I-. A solution containing Cu(I) ions is added to selectively precipitate on eof the ions. At what [Cu(I)] will a precipitate begin to form, and what is its identity? Ksp CuCl=1.0x10^-6, IKsp CuI=5.1x10^-12

Can you thoroughly explain the process of your answer, like what is the conjugate base and weak acid or work out an ICE table etc etc. Please and thank you.

Explanation / Answer

(1) HF is the weak acid and F- (in LiF) is the conjugate base

Initial moles of HF = 1.00 x 0.250 = 0.250 mol

Initial moles of LiF = 1.00 x 0.250 = 0.250 mol

Moles of LiOH added = 0.150 mol


HF + LiOH = LiF + H2O

Moles of HF = 0.250 - 0.150 = 0.1 mol

Moles of LiF = 0.250 + 0.150 = 0.4 mol


Using Henderson-Hasselbalch equation:

pH = pKa + log([LiF]/[HF])

= -log Ka + log(moles of LiF/moles of HF)

= -log(3.5 x 10^(-4)) + log(0.4/0.1)

= 4.06


(2) CuCl <=> Cu+ + Cl-

[Cl-] = 0.021 M


Ksp = [Cu+][Cl-] = [Cu+] x 0.021 = 1.0 x 10^(-6)

[Cu+] = 4.8 x 10^(-5) M when CuCl precipitates out


CuI <=> Cu+ + I-

[I-] = 0.017 M


Ksp = [Cu+][I-] = [Cu+] x 0.017 = 5.1 x 10^(-12)

[Cu+] = 3.0 x 10^(-10) M when CuI precipitates out


Thus CuI precipitates out first when [Cu+] = 3.0 x 10^(-10) M (the lower of the two concetrations)

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