Question

An 8.65-g sample of an unknown group 2a metal hydroxide is dissolved in 85.0 ml of water. an acid-base indicator is added and the resulting solution is titrated with 2.50 m hcl(aq) solution. the indicator changes color signaling that the equivalence point has been reached after 56.9 ml of the hydrochloric acid solution has been added.

Answer 1

X(OH)₂ + 2HCl = XCl₂ + 2H₂O

n{X(OH)₂}=m{X(OH)₂}/M{X(OH)₂}

n(HCl)=c(HCl)v(HCl)

n{X(OH)₂}=n(HCl)/2


m{X(OH)₂}/M{X(OH)₂}=c(HCl)v(HCl)/2

M{X(OH)₂}=2m{X(OH)₂}/{c(HCl)v(HCl)}

M{X(OH)₂}=2*8.65/{2.50*56.9*10⁻³}=121.6 g/mol

M(OH)=17.0 g/mol
M(X)=121.6-17.0*2=87.6 g/mol ⇒ X=Sr,  strontium

Sr(OH)₂ + 2HCl = SrCl₂ + 2H₂O

Answer 2

Answer;

The molar mass of metal hydroxide = 121.66 g/ mole

The formula is Sr(OH)2

Explanation;

M(OH)2 + 2 HCl = MCl2 + 2 H2O  

Moles of the acid = 2.5 x 56.9 / 1000 = 0.14225  

Therefore; moles of hydroxide at the eq point = 0.0711 moles  

These have a mass of 8.65g

Thus; mass of 1 mole = 8.65 / 0.0711

         = 121.66 g / mole  

That means the compound is Sr (OH)2 ..