Six problems. Here is how you do the first one (NH4Cl). The pH of the solution is determinded by the hydrolysis of the NH4^+. Newer texts say the NH4^+ is a weak acid. (The Cl^- is a strong base and isn't hydrolyzed).
............NH4^+ + H2O ==> NH3 + H3O^+
I..........0.25M........................0.............0
C.............-x...........................x.............x
E............0.25-x..................... x.............x
Ka(for NH4^+) = Kw/Kb(for NH3) = (NH3)(H3O^+)/(NH4^+)
Kw is 1E-14; Kb for NH3 is 1.75E-5
(1E-14/1.75E-5) = (x)(x)/(0.25-x)
Solve for x = (H3O)^+ and convert to pH.
Post your work on any of these if you get stuck. Don't use my values of Ka and Kw but look these values up in your text.
For weak acids that hydrolyze, such as the OCl^- of NaOCl, those equations are similar to the following.
..............OCl^- + HOH ==> HOCl + OH^-
When you solve for x you will see that equals (OH^-) so you convert that to pOH, then use Kw = pH + pOH to find pH.
List the following in order of increasing pH of their 0.25 M solutions: NH4Cl, MgCl2, HCl, NaOCl, C6H5NH3Cl and NaClO2.
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