Question

How many milliliters of .100 M NAOH are required to neutralize the following solutions20.0mL of .126 M HNO3 =252?75.5mL of .215 M H2SO4 = ?

Answer 1

So,

The balanced chemical reaction for this neutralization reaction (The first one) is:

`NaOH_((aq))+HNO_(3(aq))\to NaNO_(3(aq))+H_2O_((l))`

So,

We need to first determine the number of moles of HNO3 from its volume and molarity. Then convert it to moles of NaOH using stoichiometry. Finally, use the calculated moles of NaOH and its molarity (0.100 M) to figure out the volume needed.

To determine the number of moles of HNO3, we could use the formula for molarity:

`M=(n)/(V)`

Where n is the number of moles of the solute (In this case, HNO3), and V is the volume of the solution. First, convert 20.0mL of HNO3 to L:

`20.0mlHNO_3=0.02L`

Now,

`\begin{gathered} n=M\cdot V \\ n=0.126M\cdot0.02L \\ n=0.00252moles \end{gathered}`

Moles of NaOH:

Using the stoichiometry of the reaction, we got that:

`0.00252molesHNO_3\cdot\frac{1\text{molNaOH}}{1molHNO_3}=0.00252molesNaOH`

Finally, using the molarity of NaOH:

`\begin{gathered} 0.100M=(0.00252molesNaOH)/(L) \\ \\ L=(0.00252molesNaOH)/(0.100M)=0.00252L \end{gathered}`

If we pass this to mL, we got:

0.00252L NaOH = 252mL.

Therefore, 252 mL of NaOH are required to neutralize the solution.