The buffer solution is used to control the ph to insure that it does not become too high because excessively basic solutions could cause the corresponding hydroxides of hard metal ions (such as ca(oh)2 and mg(oh)2) to precipitate. using the calcium ion as a typical representative, just how high a ph do you think could be considered as "too high" for a solution with a hardness of about 250 ppm caco3? ksp for ca(oh)2 is 6.5 x 10-6.

pH = 12.7

Explanation:

First, we have to calculate the [Ca²⁺] in a solution of about 250 ppm CaCO₃.

Now, let's consider the dissolution of Ca(OH)₂ in water.

Ca(OH)₂(s) ⇄ Ca²⁺(aq) + 2 OH⁻(aq)

The solubility product Ksp is:

Ksp = [Ca²⁺] × [OH⁻]²

[OH⁻] = √(Ksp/[Ca²⁺]) = √(6.5 × 10⁻⁶/2.5 × 10⁻³) = 5.1 × 10⁻² M

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (5.1 × 10⁻²) = 1.3

pH + pOH = 14 ⇒ pH = 14 - pOH = 14 - 1.3 = 12.7

c, liquids evaporate, creating a gas

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole