Std 10 gujarati ch 19 swadhyay solution , dhoran 10 gujarati ch 19 ek bapore svadhayay solution

Calculate the pH at equivalence point when a solution of 0.10 M acetic acid is titrated with a solution of 0.10 M hydroxide (K_(a) " for acetic acid is " 1.9 xx10^(-5))

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)

Calculate OH^- concentration at the equivalent point when a solution of 0.1 M acetic acid is titrated with a solution of 0.1 M NaOH. Ka for the acid = 1.9 xx 10^(-5) .

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 solution of the equation x + 15 = 19 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)

Which of the following solutions when added to 1L of a 0.1 M CH_3COOH or in the pH of the solution K_a=1.8xx10^(-5) for CH_3COOH ?

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 ).

An aqueous solution of CH_3 COONa is