# Calculating concentration of OH⁻ ions [closed]

Calculate concentration of $$\ce{OH-}$$ ions in $$\pu{0.66 mol L^-1}$$ solution of $$\ce{NH4+}.$$

I think that there might be something missing, as the only law I have for this kind of problems is

$$[\ce{OH-}]^2 = K_\mathrm{b} × C_\mathrm{b},$$

but is there a way to calculate $$[\ce{OH-}]$$ by using concentration of base only $$(C_\mathrm{b} = \pu{0.66 mol L^-1}).$$

• I don't understand the question either, there is no such thing as solution of $\ce{NH4+}.$ Was it probably translated from another language? If it's ammonium hydroxide solution, then the answer is too obvious: $[\ce{NH4+}] = [\ce{OH-}],$ which would be strange. – andselisk May 30 at 12:12
• Yeah it probably means ammonium hydroxide. – John May 30 at 12:15
• It was a multiple choice question and 0.66 M wasn't one of the choices, so I suggest that this is the concentration of the base itself not the ions. – John May 30 at 12:18
• And since ammonium hydroxide is a weak electrolyte, then concentration of the base must be greater than concentration of hydroxide ions. – John May 30 at 12:19
• @andselisk One simply cannot have that much ammonium cation in simple NH3 solution. That means is an ammonium salt, for example NH4Cl – Mithoron May 30 at 14:11

Your problem seems to have no solution without further information. You may be interested in the the general method for any acid /base calculation is summarised below.

When there is solution of a weak acid and it salt, or just the weak acid or just the salt, i.e. pure HA, NaA + HA or pure NaA, then there is no distinction between these types of solution because of the equilibria involved. ( NaA just represents any salt ).

The following calculations apply for each or any of these solutions.

The full calculation for problems of this type starts by defining the equilibrium constants for the reactions. A molecule dissociates as $$\mathrm{HA \leftrightharpoons H^+ + A^-}$$, but to be general we let the base be [B] instead of $$\mathrm{[A^-]}$$ and so write $$\mathrm{HA \leftrightharpoons H^+ + B}$$

$$K_a=\mathrm{[H^+]_e[B]_e/[HA]_e} \tag{1}$$

where the concentrations are those at equilibrium.

There is also the equilibrium $$\mathrm{H_2O \leftrightharpoons H^+ + OH^-}$$ to consider and

$$K_w=\mathrm{[H^+]_e[OH^-]_e}$$

(The activity of water is 1 so is ignored in this equilibrium ).

We know the amounts of acid $$c_a$$ and base $$c_b$$ added at the start of the reaction. To obtain the pH at equilibrium the amount $$\mathrm{[H^+]}$$ has to be calculated. To do this work out the concentration of acid and base in terms of the initial amount and the ionized species.

The total concentration of HA is

$$c_a=\mathrm{[HA]_e+[H^+]_e - [OH^-]_e}\tag{1a}$$

and for the base

$$c_b=\mathrm{[B]_e -[H^+]_e + [OH^-]_e\tag{1b}}$$

The e subscripts are now dropped for clarity. (Note that some authors use mass and charge balance instead to work out the concentrations.)

These values can be substituted into the equilibrium constant equation (1) and using $$K_w=\mathrm{[H^+][OH^-]}$$. Substitution leads to the general equation that can be used to solve all problems of this type;

$$K_A=\mathrm{[H^+]}\frac{c_b + \mathrm{[H^+]} - K_w/\mathrm{[H^+]} }{c_a - \mathrm{[H^+]} + K_w/\mathrm{[H^+]} }\tag{2}$$

which is a cubic equation in $$\mathrm{[H^+]}$$ that is best solved numerically in the most general case. This equation can be simplified under different conditions.

Rewriting eqn 2 gives

$$K_A=\mathrm{[H^+]}\frac{c_b + \mathrm{[H^+]} - \mathrm{[OH^-]} }{c_a - \mathrm{[H^+]} + \mathrm{[OH^-]} }$$

assuming that $$\mathrm{[OH^-]} \gg \mathrm{[H^+]}$$ then

$$K_A=\mathrm{[H^+]}\frac{c_b - \mathrm{[OH^-]} }{ \mathrm{[OH^-]} }$$

As $$K_w=K_AK_B$$ then

$$K_B = \frac{\mathrm{[OH^-]^2}}{c_B-\mathrm{[OH^-]}}$$

and making a further approximation, if $$c_b >\mathrm{[OH]^-}$$ then $$K_Bc_B\approx \mathrm{[OH^-]^2}$$, which is the equation you quote. Therefore there seems that your problem cannot be solved without more information.