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

  • $\begingroup$ You need to calculate the contribution of the HCl to the pH. That will be far more important than the ammonium chloride $\endgroup$ – Michael Lautman Jun 11 '19 at 13:27
  • $\begingroup$ the contribution of hcl is already in balance $\endgroup$ – Abruchi Jun 11 '19 at 13:33
  • $\begingroup$ 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/… $\endgroup$ – porphyrin Jun 11 '19 at 16:01
  • 1
    $\begingroup$ 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. $\endgroup$ – Poutnik Jun 12 '19 at 6:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.