Question

How many milliliters of 1.72 M KOH should be added to 100. mL of solution containing 10.0 g of histidine hydrochloride (His·HCl, FM 191.62) to get a pH of 9.30? pKa=9.28

Answer 1

Given :- pH = 9.3 , hence p(OH) = 14 - 9.3 = 4.7 & pK_a  = 9.28 , hence pK_b   = 14 - 9.28 = 4.72

Apply Henderson's equation for base , ie.

p( OH ) = pK_b + log{ [salt]  / [base]

4.7 = 4.72 + log {0.52* / [ base ]}

4.7= 4.72 + log 0.52 - log [base]

4.7 = 4.72 - 0.2840 - log [ base ]

log [base ] = - 0.264

therefore, [ base ] = antilog of (-0.264) = 0.5445M

Now calculate the volume of 1.72M KOH required to be added to 100ml of 0,5445M base solution using relation-

M _KOH  x V_KOH  = M_base  x V_base

1.72 x V_KOH   = 0.5445 x 100

V_KOH   = 31.65 ml.

Thus Volume of 1.72M KOH required to be mixed = 31.65 mililitres

{ *Note that (10/191.62) = 0.052moles of salt are present in 100 ml of solution, which is equivalent to 0.52 moles are present in 1000ml. of solution. So the salt concentration would be 0.52 moles/ litre , thus [ salt ] =0.52M

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