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In: Chemistry

An aqueous solution contains 0.341 M nitrous acid. How many mL of 0.393 M potassium hydroxide...

An aqueous solution contains 0.341 M nitrous acid. How many mL of 0.393 M potassium hydroxide would have to be added to 150 mL of this solution in order to prepare a buffer with a pH of 3.140?

Expert Solution

no of moles of HNO2 = molarity * volume in L

=0.341*0.15 = 0.05115moles

no of moles of KOH = molarity * volume in L

= 0.393*v

HNO2(aq) + KOH(aq) ---------> KNO2(aq) + H2O(l)

I 0.05115 0.393*v 0

C -0.393*v -0.393*v 0.393*v

E 0.05115-0.393*v 0 0.393*v

PH = 3.14

PKa = 3.15

PH = PKa + log[KNO2]/[HNO2]

3.14 = 3.15 + log0.393*v/( 0.05115-0.393*v)

log0.393*v/( 0.05115-0.393*v) = 3.14-3.15

log0.393*v/( 0.05115-0.393*v) = -0.01

0.393*v/( 0.05115-0.393*v) = 10^-0.01

0.393*v/( 0.05115-0.393*v) = 0.977

0.393*v = 0.977*( 0.05115-0.393*v)

v = 0.0643

volume of KOH = 0.0643L = 64.3ml >>>>answer

queen_honey_blossom answered 1 year ago

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