(1 point) A monoprotic organic acid with a Ka of 6.7 x 10-4 is 3.5 %...

Question

Question

(1 point) A monoprotic organic acid with a K_a of 6.7 x 10^-4 is 3.5 % ionized when 100.0 g of it is dissolved in 1.0 L. What is the formula weight of the acid?

science chemistry

Answer 1

Answer #1

If any acid is 3.5 % ionized it means that

Its degree of ionization $\alpha$ = 3.5 / 100 = 0.035

As $\alpha$ << 1 so,

We can use this expression :

$\alpha$ = ✓ Ka / c

So, c = Ka / $\alpha$ ² = ( 6.7 x 10^-4 / ( 0.035 )^2 ) = 0.546 M

so, as molarity is defined as the number of moles of solute dissolved per litre of the solution :

so, M = ( Mass of solute ) / [ formula weight x volume of solution (L) ]

so, formula weight of organic acid = 100 / 0.546 = 183.15 g/mol

Answer 2

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