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Daniel
Jr. Member
Jr. Member
Posts: 9
Joined: Wed Apr 11, 2007 9:26 pm

I need help

Post by Daniel »

Don't know how to solve this:
What volume of 0.128M(Molarity) HCl is needed to neutralize 2.87g of
Mg(OH)2?
Thank you.
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jaelen
Full Member
Full Member
Posts: 16
Joined: Wed Oct 03, 2007 5:03 pm
Location: Canada

Post by jaelen »

- First calculate the number of moles of Mg (OH)2 and multiply by 2. This is the number of moles of [OH]- ion

- Now, notice that this is also the amount of moles of HCl that you will need and recall that M=(n/V). If you rearrange you get:

V= n/M



The answer should be about 0.75mL
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