Question

Use this data to answer the question that follows. To reach the equivalence point, 8.2 in a titration of a weak acid and a strong base, 48.84 mL of base are required. If the pH, at 24.24 mL of base was 4.25, determine the Ka of the acid.

Answer 1

SUPPOSE A WEAK ACID IS TITRATED WITH STRONG BASE LIKE NaOH , THE PLOT IS A SHOWN.

HA+ NaOH $\Leftrightarrow$ NaA + H2O

FROM THE HENDERSON EQUATION THE VOLUME OF STRONG BASE AT WHICH CONCENTRATION OF SALT EQUALS THECONENTRATION OF THE ACID, THE CORRESPONDING pH VALUE IS SAME AS pKa OF THE ACID FROM WHICH Ka IS DETERMINED.

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FROM THE GIVEN PROBLEM THE VOLUME AT THE EQUIVALENCE POINT = 48.84ml

AT NEARLY HALF OF THE VOLUME OF THAT i.e 24.24 ML THE pH VALUE GIVEN AS = 4.25 , WHICH IS EQUAL TO pKa VALUE

SO , pKa = 4.25

-logKa = 4.25

Ka = antilog(-4.5)

= 3.16x 10^-5

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antilog of -4.5 is written as antilog (-5 + 0.5)

antilog( -5 +0.5) = antilog 0.5 x antilog-5\

= 3.16 x 10^-5