You are given a hydroxide salt and are told to determine which Group II cation it contains. You dissolve 4.30 g of the salt in 50.0 mL of water and then titrate it with hydrochloric acid (it slowly dissolves during the titration). It takes 232 mL of 0.50 M HCl to reach the endpoint. What is the salt?

CV/C'V' = n/n' where C = concentration of acid = 0.50 M = 0.50 mol/L, V = volume of acid = 232 mL = 0.232 L, C' = concentartion of hydroxide salt, V = volume of hydroxide salt = 50.0 mL = 0.050 L, n = number of moles of acid = 2 and n' = number of moles of hydroxide salt = 1

Making C' subject of the formula, we have

C' = n'CV/nV'

Substituting the values of the variables into the equation, we have

C' = n'CV/nV'

C' = 1 × 0.50 mol/L × 0.232 L/(2 × 0.050 L)

C' = 1 × 0.50 mol/L × 0.232 L/0.1 L

C' = 1 × 0.50 mol/L × 2.32

C' = 1.16 mol/L

Now, we need to find the number of moles of hydroxide salt present.

So, number of moles, n = C'V'

n = 1.16 mol/L × 0.050 L

n = 0.058 mol

Also, the number of moles n = m/M where m = mass of hydroxide salt = 4.30 g and M = molar mass of hydroxide salt.

Making M subject of the formula, we have

M = m/n

Substituting the values of the variables into the equation, we have

M = 4.30 g/0.058 mol

M = 74.1379 g/mol

M ≅ 74.14 g/mol

Since the hydroxide salt is X(OH)₂, its molar mass, M is

M = molar mass of X + 2 × molar mass of O + 2 × molar mass of H

M = x + 2 × 16 g/mol + 2 × 1 g/mol

M = x + 32 g/mol + 2 g/mol

M = x + 34 g/mol

x = M - 34 g/mol

x = 74.14 g/mol - 34 g/mol

x = 40.14 g/mol

Since x = 40.14 g/mol which is close to the molar mass of calcium, X is calcium.

So, X(OH)₂ = Ca(OH)₂

So, the salt is calcium hydroxide.

Learn more about hydroxide salt here:

https://brainly.com/question/779090