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`        If 20ml of 0.1M NaOH is added to 30ml of .0.2M ch3cooh apka=4.74,the pH of a resulting solution1)4.442)9.563)8.964)9.26`
one year ago

Arun
23521 Points
```							Number of gram equivalents(eq) = moles x acidity of a base (or bascity of an acid.)Or= Molarity x Volume in litres x acidity of a base (or bascity of an acid)So,Eq(NaOH) = 20/1000 x 0.1 x 1 = 0.002 eqEq(CH3COOH) = 30/1000 x0.2 x1 = 0.006 eqThat means acid = base. Hence a salt would be formed in the solution with total volume = 30 + 20 = 50 mlThe pH of a strong base weak acid salt solution =7 + 0.5(pKa +logC) where C is concentration of saltconcentration of salt = eq/volume =1000 x 0.005/50= 0.01SopH = 7 + 0.5(4.74 +log0.01)pH = 7 + 1.37 = 8.37
```
one year ago
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