The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

A2 Chemistry Help

Hello, would be grateful if someone could help me with this question.12.50 cm3 of a solution of hydrochloric acid (HCl) of concentration 0.20 mol dm−3 is poured into a beaker. 9.00 cm3 of a solution of the strong alkali calcium hydroxide (Ca(OH)2) of concentration 0.20 mol dm−3 is added. Calculate pH, so far I calculated the moles of acid and OH but as it is a 2:1 ratio, unsure of what to do next?
So if there is 2 moles of OH, you would have to divide it by 2 to find the moles of the HCl (moles of HCl is 1)
Reply 2
Avatar for Pre-Med101
Pre-Med101
OP
10
Original post by sakthar23
So if there is 2 moles of OH, you would have to divide it by 2 to find the moles of the HCl (moles of HCl is 1)

Okay, Thank-you
Original post by Moll.987
Hello, would be grateful if someone could help me with this question.12.50 cm3 of a solution of hydrochloric acid (HCl) of concentration 0.20 mol dm−3 is poured into a beaker. 9.00 cm3 of a solution of the strong alkali calcium hydroxide (Ca(OH)2) of concentration 0.20 mol dm−3 is added. Calculate pH, so far I calculated the moles of acid and OH but as it is a 2:1 ratio, unsure of what to do next?


I’m not too sure on this but here it goes:

Find moles of HCl 0.20 x (12.5/1000) = 0.0025

When HCl dissociates it forms one mole of H and one mole of Cl- so 0.0025/2 = 0.00125 moles of H

Concentration of H = 0.00125/(12.5/1000) = 0.1moldm-3

Then use -log10[H ]

-log10(0.1)

= pH1

Which is the correct answer for HCl I believe. I don’t really understand your question. I haven’t started the acids and bases topic yet but I believe that’s what it is.
(edited 4 years ago)
Original post by Pre-Med101
Hello, would be grateful if someone could help me with this question.12.50 cm3 of a solution of hydrochloric acid (HCl) of concentration 0.20 mol dm−3 is poured into a beaker. 9.00 cm3 of a solution of the strong alkali calcium hydroxide (Ca(OH)2) of concentration 0.20 mol dm−3 is added. Calculate pH, so far I calculated the moles of acid and OH but as it is a 2:1 ratio, unsure of what to do next?

Most of the answers here are making lots of basic mistakes, to start you need the Moles of H+ from the HCL so
0.0125 x 0.20 = 2.5 x 10^-3 moles H+ as HCl disassociates in a 1:1 ratio forming H+
Then the [OH-] from Ca(OH)2 as it neutralises H+ to form water.
0.009 x 0.2 x 2 = 3.6 x 10^-3
Then find out the left over OH- so
3.6 x 10^-3 - 2.5 x 10^-3 = 1.1 x 10^-3
You then need the [OH-] so
1.1 x 10^-3 / (total volume) 0.0215 = 0.05
You then need the Kw at 298k which is 1 x 10^-14 and using the equation Kw= [H+] x [OH-] you find out the [H+] so
1x10^-14 / 0.05 = 2 x 10^-13 = [H+]
You then -log10( 2 x 10^-13) = 12.70 PH of solution.
Also studying A levels and was set the same question, hope this helps anyone with the same issue.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you