Homework Help: High School Chemistry

Posted by Becky on Monday, March 21, 2011 at 12:58am.

10mL of 0.10M HCl is given. What is the pH? How many milliliters of 0.10M NaOH would be required to neutralize it? What is the pH of the neutralized solution? What would the pH of the solution be if you added 20mL of NaOH?

Here's what I have so far:
pH of HCl=-log(0.10M)= 1

HCl + NaOH --> NaCl + H2O
C1*V1=C2*V2
(0.10M)V1=(0.10M)(0.010L)
V1=0.010L=10mL of 0.10M NaOH

The pH of neutralized solution would be 7 because NaCl is a salt from a strong base and strong acid, and NaCl in water has a neutral pH which is 7.

Thank you in advance!

High School Chemistry - DrBob222, Monday, March 21, 2011 at 1:13am
All of these are correct. Make sure you understand the use of C1V1 = C2V2. That works ONLY when the ratio of acid to base is 1:1 as in the case of HCl and NaOH. It will not work for other cases. For example, it will not work for
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O.
If we have 20 mL H2SO4 and titrate it with 20 mL of 0.1M NaOH, then concn H2SO4 is done as follows:
moles NaOH = M x L = 0.1M x 0.02 L = 0.002 moles NaOH.
Then convert moles NaOH to moles H2SO4. From the equation, 0.002 moles NaOH = 0.002/2 = 0.001 mole H2SO4.
Then M H2SO4 = mol H2SO4/L H2SO4 = 0.001mol/0.02L = 0.05M H2SO4.

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