Hi, I need help with chemistry problems. Please answer all of them! Thanks in ad

Question

Hi, I need help with chemistry problems. Please answer all of them! Thanks in advanced.

1. The pH of a cup of coffee is 5.16. Calculate the concentration of H3O+ and OH- in the coffee.

Concentration of H3O+ = ___ M

Concentration of OH- = ____ M

2.

A) Calculate the molar concentration of OH- in a solution that contains 8.7×10-9 M H3O+.

(Enter your answer to two significant figures.)

[OH-] =

B) Calculate the molar concentration of OH- in a solution that contains 3.5×10-3 M H3O+.

(Enter your answer to two significant figures.)

[OH-] =

B) Calculate the molar concentration of OH- in a solution that contains 3.5×10-3 M H3O+.

(Enter your answer to two significant figures.)

[OH-] =

Explanation / Answer

1) Given, pH of a cup of coffee = 5.16

so, [H3O+] = 10-pH = 10-5.16 = 6.92 x 10-6 M

Again, pH + pOH = 14

=> pOH = 14 - pH = 14 - 5.16 = 8.84

Now, [OH-] = 10-pOH = 10-8.84 = 1.45 x 10-9 M

2) A) Given, [H3O+] = 8.7 x 10-9 M

so, pH = - log [H3O+] = -log ( 8.7 x 10-9) = - ( log 8.7 - 9 log 10) = -(0.94 -9) = 8.06

Again, pH + pOH = 14

=> pOH = 14 -pH = 14 -8.06 = 5.94

So, [OH-] = 10-pOH = 10-5.94 = 1.15 x 10-6 M = 1.2 x 10-6 M ( upto two significant figures)

B) Given , [H3O+] = 3.5 x 10-3 M

so, pH = - log [H3O+] = -log ( 3.5 x 10-3) = - ( log 3.5 - 3 log 10) = -(0.54 -3) = 2.46

Again, pH + pOH = 14

=> pOH = 14 -pH = 14 -2.46 = 11.54

So, [OH-] = 10-pOH = 10-11.54 = 2.88 x 10-12 M = 2.9 x 10-12 M ( upto two significant figures)

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