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

Basic

Explanation:

Considering

So,

For hydrochloric acid :

Molarity = 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 of hydrochloric acid = 0.000033 moles

For :

Molarity = 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 of = 0.000024 moles

According to the given reaction:

1 mole of reacts with 2 moles of HCl

Also,

0.000024 mole of 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 will be left over. Thus, the resulting solution will be basic.