Question

An analytical chemist determines that an estuarine water sample contains 1.5 g/liter of sulfate ion (SO4 2). What is the concentration in terms of:(a) g/L of sulfur?(b) Molar concentration of sulfate?(c) Normality?

Answer 1

Step-by-step explanation:

It is known that molar mass of
`SO^(2-)_(4)` is as follows.

Molar mass =
`32 + (4 * 16)`

= 96 g/mol

It is given that the concentration of
`SO^(2-)_(4)` is 1.5 g/L.

(a) As 96 g of
`SO^(2-)_(4)` contains 32 grams of Sulfur in 1 L. And, 1.5 g of
`SO^(2-)_(4)` contains x grams of sulfur in 1 L.

Therefore, value of x is calculated as follows.

x =
`(1.5 * 32)/(96)`

= 0.5 g/L

Hence, the concentration of sulfur is 0.5 g/L.

(b) Now, calculate the molar concentration of sulfur as follows.

Molarity =
`\frac{\text{conc. of sulfate}}{\text{molar mass of SO^(2-)_(4)}}`

=
`(1.5 g/L)/(96 g/mol)`

= 0.0156 M

Therefore, the molarity of given sample is 0.0156 M.

(c) Calculate the normality as follows.

Equivalent mass of
`SO^(2-)_(4)` =
`(96)/(2)`

= 48 g/mol

Normality =
`(conc. of SO^(2-)_(4) in g/L)/(equivalent mass of SO^(2-)_(4))`

=
`(1.5 g/L)/(48 g/mol)`

= 0.03125 N

Hence, the normality of given sample is 0.03125 N.