Titration Problem 8D - 2015 Final

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Morgan Zapasnik 1N
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Joined: Fri Jul 22, 2016 3:00 am

Titration Problem 8D - 2015 Final

Postby Morgan Zapasnik 1N » Sun Dec 04, 2016 11:58 am

Can someone please write out the chemical equation for the following problem and clarify how the titration works please?
The question is 8D from the 2015 Final, and it states:
"During a strong-base strong-acid titration calculate the pH when 49.00mL of 0.100M NaOH has been added to 50.00mL of 0.100M HCl.
And for reference, the solution is pH = 3.

Sarahi Arellano
Posts: 6
Joined: Wed Sep 21, 2016 2:59 pm

Re: Titration Problem 8D - 2015 Final

Postby Sarahi Arellano » Sun Dec 04, 2016 12:43 pm

I don't know if someone replied already, but here's what I did:
Since they are strong acids and bases, they'll go to completion, therefore the acid or base in excess with dominate the pH:
we have .005 moles HCl and .0049 moles NaOH, they react and we have .0001 moles HCl left, that HCl will react with water to completion so the moles we had of HCl will be the same moles of H3O then you find the concentration of H3O by multiplying .0001 moles H3O by .1 that's the total volume and the take the p[H30]
HCl(aq) + NaOH(aq)---> NaCl(aq) + H2O(l)
HCl(aq) + H2O(l)---> Cl-(aq)+ H3O(aq)


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