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0.1M NaOH is titrated with 0.1M HA till ...

`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 ?

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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

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]

0.1 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 HA is tritrated against 0.1 M NaOH . Find the pH the end point . Dissociation constant for the end acid HA is 5 xx 10^(-6) and degree of hydrolysis , h lt 1

When 40mL of a 0.1 MN weak base, BOH is titrated with 0.01M HC1 , the pH of the solution at the end point is 5.5 . What will be the pH if 10mL of 0.10M NaOH is added to the resulting solution ?

When a 20 mL of 0.08 M weak base BOH is titrated with 0.08 M HCl, the pH of the solution at the end point is 5. What will be the pOH if 10 mL of 0.04 M NaOH is added to the resulting solution? [Given : log 2=0.30 and log 3=0.48]

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