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### Chapter 13 Question 25

Posted: 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.

### Re: Chapter 13 Question 25

Posted: 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 $H^{+}$ of the added titrant (HCl) is equal to the moles of $OH^{-}$ 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: $HCl(aq) + NaOH(aq) \rightarrow H_{2}O(l) + NaCl(aq)$
The net ionic equation for this is:$H^{+} (aq) + OH^{-}(aq) \rightarrow H_{2}O(l)$

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 $H^{+}$ and $OH^{-}$ 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 $9.17 * 10^{-3}$ L HCl.
Hope this helps!