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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?

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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


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