Question

When 220 mL of 1.50x10^-4 M hydrochloric acid is added to 135 mL of 1.75x10^-4 M Mg(OH)2, the resulting solution will be___________.

Answer 1

Answer:

Basic

Explanation:

Considering

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

So,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

For hydrochloric acid :

Molarity = $1.50\times 10^{-4}$ M

Volume = 220 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 220×10⁻³ L

$Moles =1.50\times 10^{-4} \times {220\times 10^{-3}}\ moles$

Moles of hydrochloric acid = 0.000033 moles

For $Mg(OH)_2$ :

Molarity = $1.75\times 10^{-4}$ M

Volume = 135 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 135×10⁻³ L

$Moles =1.75\times 10^{-4} \times {135\times 10^{-3}}\ moles$

Moles of $Mg(OH)_2$ = 0.000024 moles

According to the given reaction:

$Mg(OH)_2_{(aq)}+2HCl_{(aq)}\rightarrow MgCl_2_{(s)}+2H_2O_{(aq)}$

1 mole of $Mg(OH)_2$ reacts with 2 moles of HCl

Also,

0.000024 mole of $Mg(OH)_2$ reacts with 2*0.000024 moles of HCl

Moles of HCl = 0.000048 moles

Available moles of HCl = 0.000033 moles

Limiting reagent is the one which is present in small amount. Thus, HCl is limiting reagent.

Thus, HCL will be consumed completely and $Mg(OH)_2$ will be left over. Thus, the resulting solution will be basic.