Home
Class 12
CHEMISTRY
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 ?

Text Solution

Verified by Experts

9
Promotional Banner

Similar Questions

Explore conceptually related problems

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

Calculate the pH of 1.0xx10^(3) M solution of NaCI

When 40mL of a 0.1 M weak base, BOH is titrated with 0.01M HCl , 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]

15mL of sample of 0.15 M NH_(3)(aq) is titrated against 0.1 M HCl(aq) What is the pH at the end point? K_(b) of NH_(3)(aq)=1.8xx10^(-5) .Report the answer rounding it off to the nearest whole number.

Calculate the hydrolysis constant of NH_4Cl . Determine the degree of hydrolysis of this salt in 0.01 M solution and the pH of the solution. Kb( NH_4OH )=1.8× 10^(−5)

In the equilibrium A^(-)+ H_(2)O hArr HA + OH^(-) (K_(a) = 1.0 xx 10^(-4)) . The degree of hydrolysis of 0.01 M solution of the salt is

10 mL of H_(2)A (weak diprotic acid) solution is titrated against 0.1M NaOH. pH of the solution is plotted against volume of strong base added and following observation is made. If pH of the solution at first equivalence point is pH_(1) and at second equivalence point is pH_(2^.) Calculate the value of (pH_(2)-pH_(1)) at 25^(@)C Given for H_(2)A,pK_(a_1) =4.6 and pK_(a_2) =8, log 25=1.4