PH Question?

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1 Answer
Sep 19, 2016

Answer:

(A)

pH = 4.26

(B)

pH = 9.7

V = 1 L

Explanation:

(A)

A quick way of finding the pH of a weak acid is to use the expression:

#sf(pH=1/2(pK_a-loga))#

#sf(pK_a=-logK_a=-log(3xx10^(-8))=7.523)#

#sf(a)# is the concentration of the acid.

#:.##sf(pH=1/2[7.523-(-1)]=4.26)#

(B)

You need to do the volume first:

The equation is:

#sf(HClO+KOHrarrKClO+H_2O)#

#sf(n_(HClO)=cxxv=0.1000xx100/1000=0.01)#

Since they react in a 1:1 molar ratio, the number of moles of base must be the same:

#sf(n_(KOH)=0.01)#

#sf(c=n/v)#

#:.##sf(v=n/c=0.01/0.01=1color(white)(x)L)#

This is a bad question as you would never set up a titration which requires 1L for an end-point.

#sf(ClO^-)# is the co - base and is hydrolysed by water:

#sf(ClO^(-)+H_2OrightleftharpoonsHClO+OH^-)#

#sf(K_(b)=([HClO][OH^(-)])/([ClO^(-)])#

These are equilibrium concentrations.

We can use this expression to find #sf([OH^-])# hence #sf([H^+])# and the pH.

We can find #sf(K_b)# using the expression:

#sf(K_bxxK_a=K_w=10^(-14)color(white)(x)"mol"^2."l"^(-2))# at #sf(25^@C)#

#:.##sf(K_b=10^(-14)/(3.0xx10^(-8))=0.333xx10^(-6)color(white)(x)"mol/l")#

Now we can set up an ICE table based on concentrations in mol/l:

The total volume = 100 ml + 1000 ml = 1.1 L

#:.# #sf([ClO^-]=n/v=0.01/1.1=0.00909color(white)(x)"mol/l")#

#" "##sf(ClO^(-)" "+" "H_2O" "rightleftharpoons" "HClO" "+" "OH^-)#

#sf(color(red)(I)" "0.00909" "0" "0)#

#sf(color(red)(C)" "-x" "+x" "+x)#

#sf(color(red)(E)" "(0.00909-x)" "x" "x)#

#:.##sf(K_b=x^2/((0.00909-x))=0.333xx10^(-6)color(white)(x)"mol/l")#

Because the dissociation is so small I will make the assumption that #sf((0.00909-x)rArr0.00909)#.

#:.##sf(x^2=0.333xx10^(-6)xx0.00909=0.0032xx10^(-6))#

#:.##sf(x=sqrt(0.0032xx10^(-6))=5.5xx10^(-5)color(white)(x)"mol/l")#

This is equal to #sf([OH^-])#.

The ionic product of water is given by:

#sf(K_w=[H^+][OH^-]=10^(-14)color(white)(x)"mol"^2."l"^-2)# at #sf(25^@C)#

#:.##sf([H^+]=10^(-14)/[[OH^-]]=10^(-14)/(5.5xx10^(-5))=1.818xx10^(-10)color(white)(x)"mol/l")#

#sf(pH=-log[H^+]=-log[1.818xx10^(-10)]=9.7)#

This is the salt of a weak acid and a strong base so you can see that, at the equivalence point, the solution is slightly alkaline.