Determine the equilibrium pH for .0001 M Ammonium Chloride, NH4Cl (salt of the w

Question

Determine the equilibrium pH for .0001 M Ammonium Chloride, NH4Cl (salt of the weak conjugate acid, pK=9.25).

How much hypochlorous acid (HOCl; pKa=7.60) must be added to pure water to make solution of pH 4.3? pH 6.5?

The method my instructor uses is to set up equations for equilibria, mass balance, and charge balance. Then making assumptions, solve for [H+] using a system of equations but I'm not sure about the assumptions. If some one could answer these questions/explain this method I would greatly appreciate it!

Explanation / Answer

NH4Cl -----> NH3 + H+ + Cl-

0.0001-X ..........X........X

pK = 9.25

=> K = 5.623 x 10^-10

K = 5.623 x 10^-10 = X^2 / 0.0001 - X

=> X = 2.37 x 10^-7 M = pH+]

pH = - log [H+] = 6.625

pK = 7.6

=> K = 2.512 x 10^-8

HOCl -----> H+ + OCl-

C-X .............X.......X

pH = 4.3

=> [H+] = 5.012 x 10^-5 M = X

X^2 / C-X = K = 2.512 x 10^-8

=> 2.512 x 10^-9 / C - 5.012 x 10^-5 = 2.512 x 10^-8

=> C - 5.012 x 10^-5 = 0.1

=> C = 0.1 M

Therefore the HOCl solution should be made of 0.1 Molar for pH = 4.3

pH = 6.5

=> [H+] = 3.162 x 10^-7 M

X^2 / C-X = 2.512 x 10^-8

=> 10^-13 / C - 3.162 x 10^-7 = 2.512 x 10^-8

=> C = 4.3 x 10^-6 M of HOCl solution

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