Question

How many mL of 0.50 M NaOH are required to neutralize 100 mL 1.2 M HCl?Select one:a.60 mLb.120 mLc.30 mLd.240 mL

Answer 1

`D\text{ : 240 mL}`

Step-by-step explanation:

We start by writing the equation of the reaction

We have this as:

`NaOH_((aq))\text{ + HCl}_((aq))\text{ }\rightarrow\text{ NaCl}_((aq))\text{ + H}_2O_((l))`

Now, we proceed to write the equation for standardization. We have this as:

`(C_aV_a)/(C_bV_b)\text{ = }(n_a)/(n_b)`

where:

Ca is the molarity of HCl which is 1.2M

Cb is the molarity of NaOH which is 0.50M

Va is the volume of the acid which is 100mL

Vb is the volume of the base which is what we want to calculate

na is the number of moles of the acid which is 1 (as seen from the balanced equation of neutralization)

nb is the number of moles of the base which is 1

Substituting the values, we have it that:

`\begin{gathered} (1.2*100)/(0.5* V_b)\text{ = }(1)/(1) \\ \\ V_b\text{ = }(120)/(0.5)\text{ = 240 mL} \end{gathered}`