Question

Find the concentration of h3o+ aq in a 1.75 m solution of lactic acid, hc3h5o2 at 25 c ka= 6.4x10^-5

Answer 1

Hello!

The chemical equation for the dissociation reaction is the following:

CH₃CH(OH)COOH + H₂O ⇄ CH₃CH(OH)COO⁻ + H₃O⁺

The expression for the equilibrium constant Ka is the following:


`Ka=([CH_3CH(OH)COO^(-)][H_3O^(+)])/([CH_3CH(OH)COOH])`

For calculating the concentration of H₃O⁺ we need to clear from this last equation:


`Ka=([CH_3CH(OH)COO^(-)][H_3O^(+)])/([CH_3CH(OH)COOH]) \\ \\ Ka=([x][x])/([1,75M-x]) ---(1,75M-x\approx 1,75M) \\ \\ \sqrt{6,4*10^(-5)*(1,75M)}=x=0,0106M=[ H_3O^(+)]`

So, the concentration of H₃O⁺ will be 0,0106 M

Have a nice day!