Question

What is the molarity of a NaOH solution that required 27.6 mL to neutralize 30.0 mL of a 0.175 M HCL solution? Using HCl + NaOH arrow NaDCl + H2O

Answer 1

`0.190\text{ M}`

Step-by-step explanation:

Here, we want to get the molarity of the NaOH solution

We start by writing the balanced equation of reaction:

`HCl_((aq))\text{ + NaOH}_((aq))\text{ }\rightarrow\text{ NaCl}_((aq))\text{ + H}_2O_((l))`

The mathematical formula to use here is that:

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

where:

Ca is the molarity of the acid which is 0.175M

Cb is the molarity of the base which we want to calculate

Va is the volume of the acid which is 30 mL

Vb is the volume of the base which is 27.6 mL

na is the number of moles of acid in the balanced equation which is 1

nb is the number of moles of base in the balanced equation which is also 1

Substituting the values, we have it that:

`\begin{gathered} (0.175*30)/(C_b*27.6)=\text{ }(1)/(1) \\ \\ C_b\text{ = }(0.175*30)/(27.6)\text{ = 0.190 M} \end{gathered}`