`0.1M NaOH` is titrated with `0.1M, 20mL HA` till the point. `K_(a)(HA) = 6 xx10^(-6)` and degree of dissociation of `HA` is neglible (small) as compared to unity. Calculate the `pH` of the resulting solution at the end point [Use `log 6 ~~ 0.8]`

0.1M NaOH is titrated with 0.1M HA till the end point. K_(a) of HA is 5.6xx10^(-6) and degree of hydrolysis is less compared to 1. Calculate pH of the resulting solution at the end point ?

0.1M NaOH is titrated with 0.1M HA till the end point. K_(a) of HA is 5.6xx10^(-6) and degree of hydrolysis is less compared to 1. Calculate pH of the resulting solution at the end point ?

0.1 M of HA is titrated with 0.1 M NaOH, calculate the pH at end point. Given K_(a)(HA)=5xx10^(-5) and alpha lt lt 1

0.1 M of HA is titrated with 0.1 M NaOH, calculate the pH at end point. Given K_(a)(HA)=5xx10^(-5) and alpha lt lt 1

When a salt reacts with water resulting into formation of acidic or basic solution, the process is referred to as salt hydrolysis. The pH of salt solution can be calculated using the following equations. pH = ( 1)/(2) ( p K_(w) + pK_(a) + log C) for salt of weak acid and strong base. pH = (1)/(2) ( pK_(w)- pK_(b) - log C ) for salt of weak base and strong acid. pH = (1)/(2) ( pK_(w) + pK_(a) - pK_(b)) for salt of weak acid and weak base equal volume of 0.1M solution of weak acid HA is titrated with 0.1M NaOH solution till the end point pK_(a) for acid is 6 and degree of hydroglysis is less compared to 1.The pH of the resultant solution at the end point is

A monoprotic acid in 0.1 M solution has K_(a)=1.0xx10^(-5) . The degree of dissociation for acid is

50 mL of 0.1 M HCl and 50 mL of 2.0 M NaOH are mixed. The pH of the resulting solution is

0.1M solution of which of the following has almost unity degree of dissociation ?