If it takes 720. mL of 0.00125 M Mg(OH)2 to neutralize 425 mL of an HCI solution, what is the concentration of the HCI? Mg(OH)2 + HCl →MgCl2
+ H20

0.004235 M

Explanation:

The neutralization reaction of this question is given as follows:

Mg(OH)2 + 2HCl → MgCl2 + 2H2O

Using the formula below:

CaVa/CbVb = na/nb

Where;

Ca = concentration of acid (M)

Cb = concentration of base (M)

Va = volume of acid (mL)

Vb = volume of base (mL)

na = number of moles of acid = 2

nb = number of moles of base = 1

From the information provided in this question;

Ca = ?

Cb = 0.00125 M

Va = 425 mL

Vb = 720 mL

na (HCl) = 2

nb = (Mg(OH)2) = 1

Using the formula; CaVa/CbVb = na/nb

Ca × 425/0.00125 × 720 = 2/1

425Ca/0.9 = 2

472.2Ca = 2

Ca = 2/472.2

Ca = 0.004235 M

4

explanation:

it's all four options.