# Calculate the pH

Calculate the pH of a dissolution of 100 ml of 0.1 M solution of ammonium chloride in the style of 100 ml of 0.1 M hydrochloric acid solution

$$\ce{NH4Cl -> NH4+ + Cl-}$$

$$\ce{HCl -> H+ + Cl-}$$

$$\dfrac{(\pu{0.1M})(\pu{0.1L})}{\pu{0.2L}} = \pu{0.05M}$$

NH4+   +   H20 <-> NH3 + H3O+

[]   0.05        -       -    0.05

eq[] 0.05+x      x       x    0.05+x

$$K_\mathrm{b} = \pu{1.8*10^{-2}}$$

$$K_\mathrm{b} = \dfrac{(x(0.05+x))}{(0.05-x)}$$

From this equation I get $$x = \pu{1.799*10^{-5}}$$

$$\ce{pOH = [H3O+] = 1.3}$$

$$\ce{pH} = 12.7$$

Is this solved right?

Sorry but I don't know how to put it pretty

• You need to calculate the contribution of the HCl to the pH. That will be far more important than the ammonium chloride – Michael Lautman Jun 11 '19 at 13:27
• the contribution of hcl is already in balance – Abruchi Jun 11 '19 at 13:33
• The NH4 undergoes hydrolysis so an acidic solution results. The $K_b=1.6\cdot 10^{-5}$ for NH3 which gives a pH of $\approx 5.1$. ( Is there a typo in your value?) There is a general method see, chemistry.stackexchange.com/questions/60068/… – porphyrin Jun 11 '19 at 16:01
• The NH4Cl is just distraction. pH of HCl itself is so low that hydrolysis of NH4+ is fully negligible, considering implicit simplification due activity versus concentration. As [NH3] Is by about 8 orders lower than [NH4+], so does [H+] due this hydrolysis. Note that correction for activity by the simplest, Debye-Hueckel equation, is about +0.16 pH. Real one will be somewhat lower, but comparable. – Poutnik Jun 12 '19 at 6:59