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How many moles of NaOH are needed to neutralize the acid?

Asked Oct 22, 2009, 08:53 AM — 17 Answers
A 38mL solution 0.026M solution of HCl reacts with a 0.032M NaOH.

HCl+NaOH=NaCl+H2O
number of moles of HCl originally present is 0.000988moles.

how many moles of NaOH are needed to neutralize the acid?
Please show me how to do this and , Please give me answer. Please..

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17 Answers
sarah1004's Avatar
sarah1004 Posts: 107, Reputation: 1
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#11

Oct 22, 2009, 11:07 AM
31250/0.000988=3.16x10^7

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Unknown008's Avatar
Unknown008 Posts: 8,147, Reputation: 3745
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#12

Oct 22, 2009, 11:12 AM
Nope. Look back at my example. How did I get the volume of HCl for 0.23 moles?
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sarah1004's Avatar
sarah1004 Posts: 107, Reputation: 1
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#13

Oct 22, 2009, 11:15 AM
1000/0.000988=1.01x10^6

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Thank you!
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Unknown008's Avatar
Unknown008 Posts: 8,147, Reputation: 3745
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#14

Oct 22, 2009, 11:20 AM
No again.

The answer is:

0.032 moles -> 1000 mL
1 mole -> 1000/0.032 = 31250 mL
0.000988 mol -> 31250*0.000988 = 30.875 mL

Just in case, '*' means multiplication.
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sarah1004's Avatar
sarah1004 Posts: 107, Reputation: 1
Junior Member
  
#15

Oct 22, 2009, 11:27 AM
Oh.....
Now I got it thank you so much....

So, we can conclude like it

A 38mL solution 0.026M solution of HCl reacts with a 0.032M NaOH.

A)write a balanced chemical reaction for the above process.
HCl+NaOH=NaCl+H2O

B)calculate the number of moles of HCl originally present.
(0.026moles/Liter)*(38mL/1)*(1Liter/1000mL)=0.000988moles.

C)how many moles of NaOH are needed to neutralize the acid?
(0.026/1000)*38=0.000988mloles

D)calculate the volume of the NaOH required to react completely with the acid.
31250*0.000988=30.875mL


Can you check it out for me?
Thank you sososo much
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Unknown008's Avatar
Unknown008 Posts: 8,147, Reputation: 3745
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#16

Oct 22, 2009, 11:28 AM
Well done!
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sarah1004's Avatar
sarah1004 Posts: 107, Reputation: 1
Junior Member
  
#17

Oct 22, 2009, 11:31 AM
Thank you
Are you a teacher?
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Unknown008's Avatar
Unknown008 Posts: 8,147, Reputation: 3745
Uber Member
  
#18

Oct 22, 2009, 10:24 PM
Nope, a student. Perhaps. At a slightly higher level than you...
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