Q3. A third beaker, C, contains 100.0 cm^3 of 0.0125 mol/dm^3 ethanoic acid ( Ka = 1.74 × 10^−5 mol/dm^3 at 25 ºC). Write an expression for Ka and use it to calculate the pH of the ethanoic acid solution in beaker C.

A3.   Ethanoic acid is a weak acid with the formula CH3COOH. It will dissociate into H+ and CH3COO- therefore the expression for Ka is [H+][CH3COO-]/[CH3COOH]   The [H+] = [CH3COO-] and so the expression becomes, Ka = [H+]^2 / [CH3COOH]   We know the Ka value and the [CH3COOH] from the question and so can rearrange the expression to find [H+] and hence the pH   [H+] = sqrt(Ka x [CH3COOH])    Ka = 1.74 x 10^-5 mol/dm^3, [CH3COOH] = 0.0125 mol/dm^3 so [H+] = 4.66 x 10-4   pH = -log[H+] = 3.33

Answered by Tutor51285 D. Chemistry tutor

7988 Views

See similar Chemistry A Level tutors

Related Chemistry A Level answers

All answers ▸

In the topic of transition metals, what are the different types of ligands and what in itself, is a ligand?


Magnesium is in Group 2 of the Periodic Table. It has a number of naturally occurring isotopes, including 24Mg and 26Mg. (a) (i) Explain, in terms of the subatomic particles in the atoms, why 24Mg and 26Mg are isotopes.


what is the number of ion in 7.41g of calcium hydoxide Ca(OH)2 ?


Give the electron configuration of fluorine


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy
Cookie Preferences