Question

i need help in completing this table. thank you so much. how will i work on the OH?

signment Saved Help Save & Exit Submit 3 attempts left Check my work Complete the following table. Acidic or Basic? [oHr] pH pOH 4.83 Basic a) 1.5 x 10- 6.66 x 1010 9 x 10 (select) 1.3 x 10-5 (select) 10 (select |( (select) 10.99 x 10lect) 9.10 d) x 10 (select)E (select) (select) e) 6.3 x 1012 (select)

Answer 1

Answer)

pH = - log[H₃O⁺], [H₃O⁺] = 10^-pH

pOH = - log[OH^-], [OH^-] = 10^-pOH

pH + pOH = pKw =14

[H₃O⁺] x [OH^-] = 10^{-14 M}

If pH > 7 then solution is basic

If pH < 7 then solution is acidic

a). [H₃O⁺] = 1.5 x 10^-5 M, pH = - log[1.5 x 10^-5 M] = 4.82

[OH^-] = 10^-14 /1.5 x 10^-5 M = 6.67 x 10^-10 M, pOH = - log[6.67 x 10^-10 M] = 9.18

Solution is acidic.

b). [H₃O⁺] = 10^-14 /1.3 x 10^-5 M = 7.69 x 10^-10 M,  pH = - log[7.69 x 10^-10 M ] = 9.11

[OH^-] = 1.3 x 10^-5 M,  pOH = - log[1.3 x 10^-5 M] = 4.89

Solution is basic.

c). pOH = 10.99,  pH = 14 - 10.99 = 3.01

[OH^-] = 10^-10.99 = 1.02 x 10^-11 M, [H₃O⁺] = 10^-3.01 = 9.77 x 10^-4 M

Solution is acidic.

d). pH = 9.10, pOH = 14 - 9.10 = 4.9

[H₃O⁺] = 10^-9.10 = 7.94 x 10^-10 M ,  [OH^-] = 10^-4.9 = 1.26 x 10^-5 M

Solution is acidic.

e). [H₃O⁺] = 10^-14 /6.3 x 10^-12 M = 1.59 x 10^-3 M,  pH = - log[1.59 x 10^-3 M ] = 2.80

[OH^-] = 6.3 x 10^-12 M,  pOH = - log[6.3 x 10^-12 M] = 11.20

Solution is acidic.

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