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8 Aug 2009, 1:44 PM
| #1 (permalink) |
| Junior Member HSC: 2009 Gender: Male
Join Date: Feb 2008
Posts: 34
Last Activity:
Yesterday, 10:40 PM  | pH question You can hide this advertisement by registering. Calculate the pH of a solution produced by mixing 50mL of 0.1mol/L HCl with 20mL of 0.05mol/L NaOH. |
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9 Aug 2009, 9:05 PM
| #2 (permalink) |
| Supreme Member HSC: 2008 Gender: Undisclosed
Join Date: May 2008
Posts: 1,278
Last Activity:
Today, 3:16 PM | Re: pH question calculate moles of each: n(HCl) = n(H+) = cV = 0.1 * 0.05 = 0.005 moles n(NaOH) = n(OH-) = cV = 0.05 * 0.02 = 0.001 moles since they react in a 1:1 ratio there will be excess H+. Quantitatively, this will be: 0.005 - 0.001 = 0.004 moles thus n(H+) = 0.004 moles after neutralisation V = 20 + 50 = 70 mL = 0.07L [H+] = n/V = 0.004/0.07 = 0.057142857 pH = -log[H+] = -log(0.057142857) = 1.243038 = 1.24 (2 dp) Similar Question was in the 2008 HSC (multiple choice). |
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