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ludmilkaskok [199]
2 years ago
8

Calculate OH- given H3 O+ = 1.40 *10^-4 M

Chemistry
1 answer:
vova2212 [387]2 years ago
8 0

3.16 X 10^-11 M is the [OH-] concentration when H3O+ = 1.40 *10^-4 M.

Explanation:

data given:

H30+= 1.40 X 10^-4 M\

Henderson Hasslebalch equation to calculate pH=

pH = -log10(H30+)

putting the values in the equation:

pH = -log 10(1.40 X 10^-4 M)

pH = 3.85

pH + pOH =14

pOH = 14 - 3.85

pOH = 10.15

The OH- concentration from the pOH by the equation:

pOH = -log10[OH-]

10.5= -log10[OH-]

[OH-] = 10^-10.5

[OH-]  = 3.16 X 10^-11 is the concentration of OH ions when hydronium ion concentration is 1.40 *10^-4 M.

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<u>Answer:</u> The correct statement is if you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would increase.

<u>Explanation:</u>

Common ion effect is defined as the effect which occurs on equilibrium when a common ion (an ion which is already present in the solution) is added to a solution. This effect generally decreases the solubility of a solute.

Equilibrium reaction of HA and KA follows the equation:

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In the equilibrium reactions, A^- ion is getting increased on the product side, so the equilibrium will shift in the direction to minimize this effect, which is in the direction of HA.

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Equilibrium constant depends on the temperature of the system. It does not have any effect on any change of pH.

pH is defined as the negative logarithm of hydrogen ions present in the solution

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As, the equilibrium is shifting in the left direction, that means concentration of H^+ ions are getting decreases. This will increase the pH of the solution.

Hence, the correct statement is if you add the soluble salt KA to a solution of HA that is at equilibrium, the pH would increase.

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