Fall 2015 Final Exam Question 8D

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Annah Khan 1B
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Fall 2015 Final Exam Question 8D

Postby Annah Khan 1B » Tue Aug 01, 2017 8:59 pm

"During a strong-base strong-acid titration calculate the pH when 49.00 mL of 0.100 M NaOH has been added to 50.00 mL of 0.100 M HCl."

The answer states:

Moles of NaOH present = (49.00/1000) x 0.1 = 4.9x10^-3
Moles of HCl present = (50.00/1000) x 0.1 = 5.0x10^-3

I was confused as to why we shouldn't convert the mL to L instead of converting the molar concentration. Could someone please explain this to me? Thanks in advance!

Chem_Mod
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Re: Fall 2015 Final Exam Question 8D

Postby Chem_Mod » Tue Aug 01, 2017 10:33 pm

The mL is converted to L, that is why the mL are divided by 1000.

Moles = volume (L) x molarity (mol.L-1)


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