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?

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Answer 1 · 2017-01-25T06:06:55

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 :)

A 0.110 M H_2SO_4H_2SO_4 solution is used to neutralize 10.0 mL of 0.085 M NaOHNaOH. What volume of the acid, in mL, is required to neutralize the base?