A solution is prepared by mixing 16ml. of 2.5M #HCl# and 40ml. of 2M #NH_4OH#. What will be the #pH# of the solution if #pK_b# is 4.75? Thank you:)

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Answer 1 · 2017-10-04T16:10:13

#sf(pH=9.25)#

Ammonium hydroxide is not a discrete substance and is better referred to as "aqueous ammonia" #sf(NH_(3(aq))#.

#sf(HCl_((aq))+NH_(3(aq))rarrNH_(4)Cl_((aq)))#

#sf(c=n/v)#

#:.##sf(n=cxxv)#

#:.##sf(n_(HCl)=2.5xx16/1000=0.04)#

#sf(n_(NH_3)=2xx40/1000=0.08)#

The #sf(NH_3)# is INXS. This means that #sf(n_(NH_4^+)=0.04)#

and the no. moles #sf(NH_3)# remaining is given by:

#sf(n_(NH_3)=0.08-0.04=0.04)#

#sf(NH_4^+)# ions undergo hydrolysis:

#sf(NH_4^+rightleftharpoonsNH_3+H^+)#

For which:

#sf(K_a=([NH_3][H^+])/([NH_4^+]))#

These are equilibrium concentrations which, because #sf(K_a)# is small, we will approximate to initial concentrations.

Rearranging:

#sf([H^+]=K_axx([NH_4^+])/([NH_3])#

We have a buffer solution here.

Because the total volume is common we can put the no. moles straight into this:

#sf([H^+]=K_axxcancel(0.04)/cancel(0.04))#

#sf([H^+]=K_a)#

#:.##sf(pH=pK_a)#

We know that #sf(pK_a+pK_b=14)#

#:.##sf(pK_a=14-pK_b=14-4.75=9.25)#

#:.##sf(pH=9.25)#