finding pH of a strong base?

Question

how to find the pH of 0.254M ammonium chloride given the Ka of NH4+ ion as 5.62x10raised to the power of -10.The answer is 4.92 but I can't figure how to work this out.

Answer 1

First off the ammonium ion is a weak acid (not a strong base). To find the pH of a weak acid or base you'll need to set up the equilibrium. For a weak acid your products with be the conjugate base of ammonium and H+.

Now set up the equilibrium constant expression for Ka. The 0.254M NH4+ is its initial concentration. (The chloride is a spectator.) Assume that the initial concentration of the products are both zero. Your change in concentration required to reach equilibrium is unknown so let it be x. The reactant loses x and the products each gain x.

You should now have expressions to represent the concentrations at equilibrium. Plug these into the equilibrium constant expression along with the value given for Ka. Solve for x. Note: you need to use the quadratic equation or simplify by assuming that x is small making the equilibrium concentration of ammonium the same as the initial concentration.

Once you have the concentration of H+ (x = [H+]), it should be fairly easy to find pH (pH = -log[H+]).

Answer 2

The standard equation is

[H+] = root (Ka x molarity).

You can assume that the equilibrium molarity of the (NH4)+ is the same as its initial molarity, because its such a weak acid.

Once you have [H+] (the concentration of hydrogen ions), you take the negative logarithm in the normal way.

Answer 3

The equivalence point would be at the point where half of the acetic acid has converted to its ionic form. For the amount you have, that would be **5**, not 10 mL of NaOH. That's your problem--you've titrated almost all of the acetic acid because you've added an equivalent number of moles of NaOH.

Answer 4

use red litmus paper