# Question 93f3b

Oct 2, 2017

$\text{100 ml}$

#### Explanation:

You have some acid ($\textcolor{b l u e}{\text{HCl}}$) and some base ($\textcolor{\mathmr{and} a n \ge}{\text{NaOH}}$) which are mixed together. What will happen? Well $\textcolor{b l u e}{\text{HCl}}$ is a strong acid and $\textcolor{\mathmr{and} a n \ge}{\text{NaOH}}$ is a strong base. What ends up happening is that they will react completely to produce a $\textcolor{p u r p \le}{\text{salt}}$ and water (color(magenta)("H"_2"O"). This is called a neutralization reaction. The reaction proceeds as so..

color(white)(aaaaaaaaaa)color(blue)"HCl"_((aq))+color(orange)"NaOH"_((aq))rarrcolor(purple)"NaCl"_((aq))+color(magenta)("H"_2"O")_[(l)]

The mole-to-mole ratio between the acid and the base is 1:1 which means $\text{1 mole}$ of $\textcolor{b l u e}{\text{HCl}}$ reacts with $\text{1 mole}$ of $\textcolor{\mathmr{and} a n \ge}{\text{NaOH}}$, based off the reaction we wrote down above.

So we first need to figure out the number of moles of $\textcolor{b l u e}{\text{HCl}}$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a} \text{C * V = moles}$

Where

C is the concentration in Molars and V is the volume in Liters. Convert ml to liters first.

• "100 ml" xx "0.1 M"-> "0.1 L" xx "0.1 M" = color(blue)("0.01 moles of HCl")

If we have $\textcolor{b l u e}{\text{0.01 moles of HCl}}$ then how much volume is required to completely neutralize the acid? Well if we have a 1:1 ratio and we have color(orange)("0.1 M of NaOH"#, then we would need $\textcolor{\mathmr{and} a n \ge}{\text{100 ml}}$ of the base to give the same number of moles.