# A. wat is the equation for the neutralization of HCl and Ca(OH)_2?

In a teaspoon (5.0mL) of a common liquid antacid, there are 190mg Ca(OH)_2} and 190mg Al(OH)_3. A 7.8×10^−2 M HCl, which is similar to stomach acid, is used to neutralize 5.0mL of the liquid antacid.

B. what is the equation for the neutralization of HCl and Al(OH)_3.?

C. What is the pH of the HCl solution?

D. How many milliliters of the HCl solution are needed to neutralize the Ca(OH)_2?

E. How many milliliters of the HCl solution are needed to neutralize the Al(OH)_3?

Im stuck=/ please i need help.

Relevance

A. Ca(OH)2 is a slightly soluble salt. It will rapidly dissolve when added to acid.

2 HCl(aq) + Ca(OH)2(aq) --> CaCl2(aq) + 2 H2O(l)

B. Al(OH)3(s) + 3 HCl(aq) --> AlCl3(aq) + 3 H2O(l)

C. pH = -log[H^+], [H^+] = [H3O^+] = 7.8 x 10^-2 M because HCl is a strong acid.

pH = -log(7.8 x 10^-2) = 1.11

D. How many milliliters of the HCl solution are needed to neutralize the Ca(OH)_2?

190 mg Ca(OH)2 = 0.190 g Ca(OH)2

moles Ca(OH)2 = 0.190 g Ca(OH)2(1 mole Ca(OH)2)/(74.093 g) = 2.56 x 10^-3 moles

From the reaction in A it takes 2 moles HCl to react with 1 mole CaCl2.

moles HCl required = (2.56 x 10^-3 moles Ca(OH)2)(2 moles HCl)/(1 mole Ca(OH)2)

moles HCl required = 5.13 x 10^-3 moles HCl

molarity HCl = (moles HCl)/(liters solution), so

liters solution = (moles HCl)/(molarity HCl)= (5.13 x 10^-3 moles HCl)/(7.8×10^−2 M HCl)

liters solution = 6.58 x 10^-2 L

mL solution = (6.58 x 10^-2 L)(1000 mL)/(1 L) = 65.8 mL of HCl needed

E. You do exactly the same thing to obtain the mL of HCl needed to neutralize 190 mg Al(OH)3 except that it takes more HCl to neutralize the Al(OH)3. I think you can do part this yourself.

• Anonymous
4 years ago

Ca Oh 2 Hcl