A 100.0 mL sample of 0.10 M #NH_3# is titrated with 0.10 M #HNO_3#. How do you determine the pH of the solution after the addition of 40.0 mL of #HNO_3#?. The Kb of NH3 is 1.8 × 10-5.?

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Answer 1 · 2016-12-29T16:06:35

You can do it like this:

As the acid is added to the base the following neutralisation takes place:

#sf(NH_(3(aq))+HNO_(3(aq))rarrNH_(4)NO_(3(aq))+H_2O_((l)))#

The initial moles of #sf(NH_3)# present is given by:

#sf(n_(NH_3)=cxxv=0.10xx100.0/1000=0.01)#

The number of moles of #sf(HNO_3)# added is given by:

#sf(n_(HNO_3)=cxxv=0.10xx40.0/1000=0.004)#

From the equation you can see that the acid and base react in a molar ratio of 1:1.

So the no. moles of #sf(NH_3)# remaining will be 0.01 - 0.004 = 0.006.

The total volume is now #sf(100.0 + 40.0 = 140.0color(white)(x)cm^3)#

The concentration of #sf(NH_3)# is given by:

#sf([NH_3]=c/v=0.006/(140.0/1000)=0.04286color(white)(x)"mol/l")#

From an ICE table we get the expression:

#sf(pOH=1/2(pK_b-logb))#

Where b is the concentration of the base.

We can approximate this to the initial concentration since the dissociation is small.

#sf(pK_b=-logK_b=-log(1.8xx10^(-5))=4.744)#

Putting in the numbers:

#sf(pOH=1/2[4.744-log(0.04286)]#

#sf(pOH=1/2[4.744-(-1.3679)]=3.056)#

At #sf(25^@color(white)(x)C)# we know that:

#sf(pH+pOH=14)#

#:.##sf(pH=14-3.056=10.9)#