Std 10 gujarati ch 17 divso judai na Jay chhe swadhyay solution , dhoran 10 gujarati ch 17 svadhayay solution

Ionisation constant of CH_(3)COOH is 1.7 xx 10^(-5) and concentration fo H^(+) in certain acetic acid solution is 3.4 xx 10^(-4) M . The concentration of acetic acid solution is

The molarity of a solution of Na_(2)O_(3) having 10.6g//500ml of solution is

20 mol of M//10 CH_(3)COOH solution is titrated with M//10 NaOH solution. After addition of 16 mL solution of NaOH . What is the pH of the solution (pK_(a) = 4.74)

0.1 M CH_(3) COOH solution is 1.0% ionized. In another diluted solution acetic acid is 10% ionised. In other solution concentration of acetic acid is

Calculate the pH at the equivalence point when a solution of 0.1 M CH_3COOH is titrated with a solution of 0.1 M NaOH. K_a(CH_3COOH)= 1.8 xx 10^(-5)

1 litre of 1M CH_(3)COOH (very weak acid) taken in a container initially. Now this solution is diluted upto volumen V (litre) so that pH of the resulting solution becomes twice the original value. (K_(a) (CH_(3)COOH) = 10^(-6)) . Now equal volume of 0.5 xx 10^(-6)M NaOH solution is added to this resulting solution, so that a buffer solution is obtained. Find pH of final solution (log 3 = 0.477 ).

If Pb[CH_(3)COO]_(2) and Na_(2)S are mixed and dissolved in water and the solution is filtered then the filterate will give test of

1 litre of 1M CH_(3)COOH (very weak acid) taken in a container initially. Now this solution is diluted upto volumen V (litre) so that pH of the resulting solution becomes twice the original value. (K_(a) (CH_(3)COOH) = 10^(-6)) . Now equal volume of 0.5 xx 10^(-6)M NaOH solution is added to this resulting solution, so that a buffer solution is obtained. Find [H^(+)] in original solution