A 0.110 M #H_2SO_4# solution is used to neutralize 10.0 mL of 0.085 M #NaOH#. What volume of the acid, in mL, is required to neutralize the base?

1 Answer
Jan 25, 2017

Answer:

You need 3.86 mL of #"H"_2"SO"_4# to neutralise the base.

Explanation:

So what we have to do is use stoichiometry to find the volume of acid needed to neutralise the base.

1: We need a balanced chemical equation:

#"H"_2"SO"_4(aq) + "2NaOH (aq)" rarr "Na"_2"SO"_4 (s) + "2H"_2"O"(l)#

So what we can find first is the moles of #"NaOH"#.

2: Solve for moles. Note that I changed 10 mL to 0.01 L.

#n_1 = cv#

# = "0.085 mol/L (0.010 L)"#

# = 8.5xx10^-4 "mol"#

3: Now we transfer the moles of #"NaOH"# to #"H"_2"SO"_4"#.

To do that, we divide by #"NaOH"#'s coefficient, and multiply by #"H"_2"SO"_4#'s.

#n_2 = (8.5xx10^-4 "mol")/2#

# = 4.25xx10^-4 "mol"#

There are #4.25xx10^-4# mol of #"H"_2"SO"_4#.

4: With #"H"_2"SO"_4#'s moles and concentration (given), we can solve for volume.

#v = n_"2"/c#

# = (4.25xx10^-4 "mol")/("0.110 mol/L")#

# "= 0.003,863,636"# L = #"3.863,636"# mL

Significant digits tells us to round the volume to 3 decimal places:

#3.86# mL.

Therefore, you need #3.86# mL of #"H"_2"SO"_4# to neutralise the base.

Hope this helps :)