Chapter 13 Question 25

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Josh Clark 1J
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Joined: Wed Sep 21, 2016 2:59 pm

Chapter 13 Question 25

Postby Josh Clark 1J » Fri Dec 02, 2016 9:11 pm

Calculate the volume of .150 M HCl(aq) required to neutralize one half of all the hydroxide ions in 25.0 mL of .110 M NaOH(aq).

Can someone please walk me through this problem. I am confused and don't know where to begin.

Britney Pheng 1L
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Joined: Wed Sep 21, 2016 2:56 pm
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Re: Chapter 13 Question 25

Postby Britney Pheng 1L » Sat Dec 03, 2016 1:32 am

Hi!
Neutralization occurs at the stoichiometric point.
For this titration, the stoichiometric point is where the moles of of the added titrant (HCl) is equal to the moles of of the analyte (NaOH).

Basically, we want to get the amount of moles of HCl = the amount of moles of NaOH.

Just to see things out, our chemical equation is:
The net ionic equation for this is:

First, we find the moles of NaOH by diving its molarity (M) by the volume (Liters).
.110 M / .025 L = 0.00275 mol (Remember to convert mL to L)

Since we want to neutralize only half of the hydroxide ions, we have to divide the moles by 2.
0.00275/2 = 0.001375 mol

Remember our moles of and must equal, so we can set our moles of HCl to 0.001375 mol NaOH, and solve for the unknown volume.
.150 M * V = 0.001375 mol (Molarity*Volume= moles)

Our answer will come out to be 0.009166 or L HCl.
Hope this helps!


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