A solution of acetic acid is $$1.0$$% ionized. Determine the molar concentration of acid Ka $$=$$ $$1.8$$ × $$10-5$$ and also the $$H+$$

Question

A solution of acetic acid is $$1.0$$% ionized. Determine the molar concentration of acid Ka $$=$$ $$1.8$$ × $$10-5$$ and also the $$H+$$

Infinity Learn · Accepted Answer

$$CH3COOHC$$ $$-$$ Cα  ⇌  $$CH3COO-C$$α  $$+$$ $$H+C$$α     where ‘C’ is the concentration of the acid and α is the degree of dissociation.$$Ka=C$$α×Cα$$C(1-$$α$$)$$ As α is very small can be neglected in the denominator $$Ka=C$$α$$2$$   Ka $$=$$ $$1.8$$ × $$10-5=$$ Cα$$2$$∴           $$1.8$$ × $$10-5=$$ $$C.0.012or$$        $$C=1.8$$ × $$10-5$$  ×  $$104$$               $$=$$   $$1.8$$ × $$10-1MH+$$ $$=$$ Cα  or  $$H+$$ $$=1.8$$ × $$10-1$$ × $$0.01$$         $$=0.18$$ × $$0.01=0.0018$$ M  or  $$1.8$$ ×  $$10-3$$  M

A solution of acetic acid is 1.0% ionized. Determine the molar concentration of acid $(K_{a} = 1.8 \times 10^{- 5})$ and also the $[H^{+}]$

Unlock the full solution & master the concept.

Ready to Test Your Skills?

free study material

A solution of acetic acid is 1.0% ionized. Determine the molar concentration of acid Ka = 1.8 × 10-5 and also the H+

Unlock the full solution & master the concept.

Ready to Test Your Skills?

free study material

A solution of acetic acid is 1.0% ionized. Determine the molar concentration of acid $(K_{a} = 1.8 \times 10^{- 5})$ and also the $[H^{+}]$