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pH question (1 Viewer)
- Thread starter Nuendo
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Pwnage101
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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).
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).