What will be the final pH on mixing 50 ml of 0.01 M methanoic acid with 50 ml of 0.01 M sodium hydroxide ?

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#pK_a# for methanoic acid is 3.75.

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Answer 1 · 2017-05-04T05:19:13

#sf(pH=8.22)#

#sf(HCOOH+NaOHrarrHCOONa+H_2O)#

#sf(n_(HCOOH)=cxxv=0.10xx50/1000=0.005)#

#:.##sf(n_(HCOONa)# which have been formed#sf(=0.005)#

From the equation you can see that the total moles of NaOH added will also be 0.005.

The volume of 0.1M NaOH added must also be 50 ml at equivalence.

#:.# Total volume = 50 + 50 = 100 ml = 0.1 L

#:.# #sf([HCOO^-]=n/v=0.005/0.1=0.05color(white)(x)"mol/l")#

You should be given the #sf(pK_a)# value for #sf(HCOOH)#.

#sf(pK_a=3.75)#

You need the #sf(pK_b)# for #sf(HCOO^-)#.

You get this from:

#sf(pK_a+pK_b=14)#

#:.##sf(pK_b=14-3.75=10.25)#

Now use this expression for the pOH of a weak base:

#sf(pOH=1/2(pK_b-log[base])#

You get this from an ICE table and assume the dissociation is negligible.

#:.##sf(pOH=1/2(10.25-log[0.05])#

#sf(pOH=1/2(10.25-(-1.3))=5.775)#

#sf(pH+pOH=14)#

#:.##sf(pH=14-pOH=14-5.775=8.22)#

#sf(HCOONa)# is the salt of a weak acid and a strong base so you would expect the pH to be slightly alkali due to salt hydrolysis:

#sf(HCOO^(-)+H_2OrarrHCOOH+OH^-)#