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?

Question

Wyatt Adkins · Accepted Answer

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

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 :)

Peyton Adams · Answer

undefined The volume of the acid required to neutralize the base is 7.85 mL.