Tutored Practice Problem 17.2.7 GOUNTS TOWARDS GRADE Prepare a buffer by acid-ba

Question

Tutored Practice Problem 17.2.7 GOUNTS TOWARDS GRADE Prepare a buffer by acid-base reactions. Close Problem Consider how to prepare a buffer solution with pH 3.14 (using one of the weak acid/conjugate base systems shown here) by combining 1.00 L of a 0.467-M solution of weak acid with 0.357 M potassium hydroxide. Weak Acid Conjugate Bae Ka HNO2 HCIO HCN NO2 CIO CNT 4.5 x 104 3.5 x 10-8 40x 1010 9.40 pKa 3.35 7.46 How many L of the potassium hydroxide solution would have to be added to the acid solution of your choice? Check & Submit Answer Show Approach

Explanation / Answer

HNO2 has pka value near to required pH of 3.14 , hence our buffer is HNO2 and NO2^-

Let KOH volume be V ( liters)

KOH moles = Molarity x volume = 0.357 V = OH- moles

HNO2 moles = 0.467 x 1 = 0.467

we have reaction HNO2 (aq) + OH- (aq) <---> NO2- (aq) + H2O (l)

HNO2 moles after reacting with 0.357 moles of OH- = 0.467 - 0.357V

NO2- moles = OH- moles reacted = 0.357V

pH of buffer = pka + log ( conjugate acid moles / acid moles)

pH = pka + log (NO2- moles / HNO2 moles)

3.14 = 3.35 + log ( 0.357V/(0.467-0.357V))

( 0.357V) / ( 0.467-0.357V = 10 ^ ( 3.14-3.35)

0.357V = 0.6166 ( 0.467-0357V)

0.577 V = 0.288

V = 0.499 L = 0.5 L ( rounded value)

Thus answer is 0.5 L

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