Question

A sample of lake water at 25°C is analyzed and the following parameters are found: total alkalinity = 6.2 x 10-4 M phenolphthalein alkalinity = 1.0 X 10PM pH = 7.6 hardness = 30.0 mg/L [Mg²+] = 1.0 x 10-4 M Extract all possible single-ion concentrations that you can by combining one or more of these data. Also determine whether or not the water is at equi- librium with respect to the carbonate-bicarbonate system and whether or not it is saturated with calcium carbonate.

Answer 1

Step-by-step explanation:

The given data is as follows.

Temperature =
`25^(o)C`, Total alkalinity =
`6.2 * 10^(-4)` M

Alkalinity of phenolphthalein =
`1 * 10^(-5)` M

pH = 7.6, Hardness = 30 mg/L

`[Mg^(2+)] = 1 * 10^(-4)` M

It is known that total alkalinity is as follows.

Total alkalinity =
`[HCO^(-)_(3)] + [CO^(2-)_(3)] + [OH^(-)] - [H^(+)]`

Since, pH = 7.6. Hence, concentration of hydrogen ions will be calculated as follows.

pH =
`-log [H^(+)]`

7.6 =
`-log [H^(+)]`

`[H^(+)]` = antilog (-7.6)

=
`2.51 * 10^(-8)` M

As, pH + pOH = 14

pOH = 14 - pH

= 14 - 7.6

= 6.4

So,
`[OH^(-)] = 10^(-6.4)`

=
`3.98 * 10^(-7) M`

Phenolphthalein alkalinity =
`[CO^(2-)_(3)]`

`[HCO^(-)_(3)]` = Total alkalinity - 2[CO^{2-}_{3}] - [OH^{-}] + [H^{+}][/tex]

=
`6.2 * 10^(-4) - 2(1 * 10^(-5)) - (3.98 * 10^(-7)) + (2.51 * 10^(-8))`

=
`5.9 * 10^(-4) M`

`[Mg^(2+)] = 1 * 10^(-4) M` =
`24 * 10^(-4) g/L`

`[Mg^(2+)]` = 2.4 mg/L

Ratio of molar mass of
`Ca^(2+)` and
`[Mg^(2+)]` with
`CaCO_(3)` is as follows.

`\frac{M_{CaCO_(3)}}{M_(Ca)}` =
`(100)/(40)` = 25

`\frac{M_{CaCO_(3)}}{M_(Mg)}` =
`(100)/(24)` = 4.2

Total hardness =
`2.5[Ca^(2+)] + 4.2 [Mg^(2+)]`

`[Ca^(2+)] = \frac{\text{total hardness - 4.2(2.4)}}{2.5}`

=
`1.99 * 10^(-4) mol/L`

`CaCO_(3) \rightarrow Ca^(2+) CO^(2-)_(3)`

`[Ca^(2+)][CO^(2-)_(3)]` =
`(1.99 * 10^(-4)) * (1 * 10^(-5))`

=
`1.99 * 10^(-9)`

`K_(sp)` of
`CaCO_(3)` =
`3.36 * 10^(-9)`

Since,
`[Ca^(2+)][CO^(2-)_(3)]` >
`K_(sp)`

Hence, the water is slightly unsaturated with
`CaCO_(3)`.