Question

A 4.36 g sample of an unknown alkali metal hydroxide is dissolved in 100.0 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 equivalent point has been reached after 17.0 ml of the hydrochloric acid solution has been added. (a) what is the molar mass of the metal hydroxide? (b) what is the identity of the metal cation?

Answer 1

(a)
`102.6g/mol`

(b) Rubidium

Step-by-step explanation:

Hello,

This titration is carried out by assuming that the volume of base doesn't have a significant change when the mass is added, thus, we state the following data a apply the down below formula to compute the molarity of the base solution:

`V_(base)=0.1L; M_(acid)=2.5M, V_(acid)=0.017L\\V_(base)M_(base)=V_(acid)M_(acid)`

Solving for the molarity of base we've got:

`M_(base)=(M_(acid)*V_(acid))/(V_(base))=(2.50M*0.017L)/(0.1L) =0.425M=0.425mol/L`

Now, we can compute the moles of the base as:

`n_(base)=0.425mol/L*0.1L=0.0425mol`

(a) Now, one divides the provided mass over the previously computed moles to get the molecular mass of the unknown base:

`(4.36g)/(0.0425mol) =102.6g/mol`

(b) Subtracting the atomic mass of oxygen and hydrogen, the metal's atomic mass turns out into:

`102.6g/mol-16g/mol-1g/mol=85.6g/mol`

So, that atomic mass dovetails to the Rubidium's atomic mass.

Best regards.