# 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

You need 3.86 mL of

#### 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

**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

To do that, we divide by

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

# = 4.25xx10^-4 "mol"#

There are

**4:** With

#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:

Therefore, you need

Hope this helps :)