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Question

Sulphurous acid (H2SO3) has Ka1 = 1.7 × 10–2 and Ka2 = 6.4 × 10–8. The pH of 0.588 M H2SO3 is _________. (Round off to the nearest integer).


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Solution

Solution:-

Step 1:- To find the value of α, by using formula Ka=ProductsReactants

Given,

Ka1=1.7*10-2Ka2=6.4*10-8

pH of solution is due to first dissociation only since Ka1>>Ka2

First dissociation of H2SO3

H2SO3(aq)H+aq)+HSO3-(aq)t=0C00t=t0.588-ααα

Ka=ProductsReactantsKa=H+HSO3-H2SO31.7100=αα0.588-α1.7*0.588-1.7α=100α2100α2+1.7α-1=0Tosolveforα,weusethequadraticformulaα=-b±b2-4ac2aα=-1.7+(1.7)2-4*100*12*100α=H+=0.0916

Step 2: Using the concentration to find pH

pH=-log[H+]pH=-log(0.0916)pH=1.0361

Final answer :- The pH of 0.588 M H2SO3 is 1.


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