Question

After diluting 30 ml of 0.100M NH4OH(Kb=1.75x10-5) to 100 ml, the solution is titrated with 0.100 M HCl. Calculate the pH value of the solution: a) at the start of the titration b) two-thirds of the way to the equivalence point c) at the equivalnece point d) when 40.0 ml of the HCl have been added. Make a graph of oH versus Va.

Answer 1

a)

Molarity of NH4OH solution on dilution to 100 ml with water = ( 30 x 0.1 ) /100

............................................................................................... = 0.03M

The equilibrium reaction is ,

NH4OH <---------------> NH4⁺   + OH^-

(0.03 - x )..............................x...................x

We, assume that [OH^- ] = x when equilibrium is reached . Thus we can write equilibrium

expression as

K_b =[NH4⁺ ] [ OH ^- ] / [ NH4OH ]

.... = x² /( 0.03 - x )

Since NH4OH is a weak base, the value of x is insignificant relative to 0.03, hence

K_b  = x² / 0.03 ; & x = SQRT {0.03 x K_{b ^}}

x = SQRT{0.03 ( 1.8 x 10^-5 )}

... = 7.35 x 10^-4 M

so [OH^- ] = 7.35 x 10^-4 M

& p(OH) = -log ( 7.35 x 10^-4 )

................ = 3.13

again , pH + p(OH) = 14

so , pH of the solution before titration = 10.87

b)

100ml of solution 0.03 M NH4OH solution yields 7.35 x 10^-4 M OH^- on dissociation

So at 2/3 of the equivalence point [ OH^- ]= 1/3 ( 7.35 x 10^-4 )

.................................................................= 2.45 x 10^-4

hence, p(OH) = -log (2.45 x 10^-4 )

....................... = 3.61

& pH ............. = 10.38   

c)

At equivalence point the pH of the solution would depend upon the hydrolysis of the salt NH4Cl

present in solution & therefore its concentration ie. [ NH4Cl ] = 2.45 x 10^-4 M

Since it is a titration between strong acid & a weak base the pH is calculated using expression,

pH = 7 - 1/2 pK_b - 1/2 log C

...... = 7 - 1/2 (4.7570 ) - 1/2 {log ( 2.45 x 10^-4 )}

....... = 7 - 2.3785 - (1.8054)

...... = 6.4269

or,....~ = 6.43

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Glad to help ; please submit 4th part (d) of the question separately as fresh question for detailed answer.