Question

ph of a solution is 6.8 calculate h+ and oh-ion in the solution
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Answer 1

Answer:

Concentration of [H^+] = 1.58 x 10^-7 M and concentration of [OH^-] is 6.32 x 10^-8 M.

Explanation:

Given: In the above question pH of a solution is given as 6.8.

To find: We need to calculate concentration of hydronium as well as hydrogen ions in the given solution.

Step Wise Solution: We know that pH of a solution is equal to negative log of hydronium ion concentration.

Therefore pH = - log[H^+] OR -log[H^+] = 6.8

So, log[H^+] = -6.8

Taking antilog we have, [H^+] = 1.58x 10^-7 M

Also we know that for a solution in water Kw = [H^+][OH^-]

Or [OH^-] = Kw/ [H^+]  (value of Kw = 1 x 10^-14)

Substituting value of Kw we have:-

[OH^-] = 1 x 10^-14/ 1.58x 10^-7 = 0.62 x 10^-7 M

OR  [OH^-] = 6.32 x 10^-8 M.

Thus, for the given solution Concentration of [H^+] = 1.58 x 10^-7 M and concentration of [OH^-] is 6.32 x 10^-8 M.

Code:SPJ2

Answer 2

Answer:

Explanation:he pH of a solution is defined as the negative logarithm of the hydrogen ion (H+) concentration in moles per liter (M):

pH = -log[H+]

So, if the pH of a solution is 6.8, the H+ ion concentration can be calculated as follows:

[H+] = 10^(-pH) = 10^(-6.8) = 7.94 x 10^(-7) M

Since the concentration of hydrogen ions and hydroxide ions are related by the equation:

Kw = [H+][OH^-] = 1.0 × 10^(-14) (at 25°C)

where Kw is the ion product constant for water, the concentration of OH- ions can be calculated as follows:

[OH^-] = Kw / [H+] = (1.0 × 10^(-14)) / (7.94 × 10^(-7)) = 1.26 × 10^(-7) M