Question

Need help with chemistry 34.0mL of sulfuric acid H_2 SO_4 is completely neutralized by 24.2 mL of 0.116 M aluminum hydroxide Al(OH)_3. What is the concentration (molarity) of the sulfuric acid?

Answer 1

`0.124\text{ M}`

Step-by-step explanation:

Here, we want to get the molarity of sulfuric acid

We start by writing the equation of reaction:

`3H_2SO_4\text{ + 2Al\lparen OH\rparen}_3\text{ }\rightarrow\text{ Al}_2(SO_4)\placeholder{⬚}_3\text{ + 6H}_2O`

Mathematically, we have it that:

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

Ca is the molarity of the acid which is unknown

Cb is the molarity of the base which is 0.116 M

Va is the volume of the acid which is 34 mL

Vb is the volume of the base which is 24.2 mL

na is the number of moles of acid in balanced reaction which is 3

nb is the number of moles of base in balanced reaction which is 2

Substituting these values:

`\begin{gathered} (C_a*34)/(0.116*24.2)\text{ = }(3)/(2) \\ \\ C_a\text{ = }\frac{0.116\text{ }*24.2*3}{2*34}\text{ = 0.124 M} \end{gathered}`