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pls Question 5 What is the pH after .03 mol of OH is added to 1.0 L of a buffer containing .05 mol of HSO, and .07 mol of so,
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Answer #1

the reaction is


HSO4-(aq) + HO- (aq) ===> SO42- (aq) + H2O(aq) :

I 0.05 0.03 0.07 0

E 0.02 - 0.10   

Since we know that

pka = 1.92

we have for buffers, pH = pKa + log [SO42- ] /[HSO4- ]

As we know wth0.03 moles of OH- added :

[SO42- ]= 0.07 moles + 0.03 moles / L = 0.10 M

[HSO4- ] = 0.05 moles - 0.03 moles / L = 0.02 M

pH = 1.92 + log (0.10/0.02)

pH =1.92+0.69

pH =  2.59

So the correct answer is option (B) 2.59

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