Apotassium hydroxide (koh) solution has a molar concentration of 0.065 m, calculate the [h3o+], [oh-] and ph of the solution.

1) [OH⁻] = 0.065 M

2) [H₃O⁺] = 1.54 x 10⁻¹³ M.

3)  pH = 12.81.

Explanation:

HCl is completely dissociated yo its ions in solution:

KOH + H₂O  → H₃O⁺ + K⁺ + OH⁻,

1) [OH⁻] = 0.065 M.

2) [H₃O⁺]:

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

∴ [H₃O⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/(0.065 M) = 1.54 x 10⁻¹³ M.

3) pH:

For strong acids like HCl:

pH = - log[H₃O⁺] = - log[1.54 x 10⁻¹³ M] = 12.81.

2b2t

explanation:

2b2t

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