Calculate the ph of each solution given the following (H3O.+) or (OH-) values Pa

Question

Calculate the ph of each solution given the following (H3O.+) or (OH-) values
Part A- (H3O+ ) = 2.5 x10-8 M Use TWO decimal places Ph=_______ Part B- (H3O+) = 6.0 x 10-6M Use TWO decimal places Part C- (OH-) = 4.0 x 10-2 M Use TWO decimal places Part D - (OH-) = 7.5 x 10-3 M Use TWO decimal places Calculate the ph of each solution given the following (H3O.+) or (OH-) values
Part A- (H3O+ ) = 2.5 x10-8 M Use TWO decimal places Ph=_______ Part B- (H3O+) = 6.0 x 10-6M Use TWO decimal places Part C- (OH-) = 4.0 x 10-2 M Use TWO decimal places Part D - (OH-) = 7.5 x 10-3 M Use TWO decimal places
Part A- (H3O+ ) = 2.5 x10-8 M Use TWO decimal places Ph=_______ Part B- (H3O+) = 6.0 x 10-6M Use TWO decimal places Part C- (OH-) = 4.0 x 10-2 M Use TWO decimal places Part D - (OH-) = 7.5 x 10-3 M Use TWO decimal places

Explanation / Answer

Part A) Given , [H3O+] = 2.5 x 10-8 M

Now, pH = - log [H3O+] = - log (2.5 x 10-8) = - (log 2.5 - 8 log 10) = - (0.40 - 8) = 7.60

Part B) Given, [H3O+] = 6.0 x 10-6 M

Now, pH = - log [H3O+] = -log (6.0 x 10-6) = -( log 6.0 - 6 log 10) = -(0.78 - 6) = 5.22

Part C) Given, [OH-] = 4.0 x 10-2 M

Now, pOH = - log [OH-] = - log (4.0 x 10-2) = -(log 4.0 - 2 log10) =- (0.60 - 2) = 1.40

Again , pH + pOH = 14 => pH = 14- pOH = 14 -1.40 = 12.60

Part D) Given, [OH-] = 7.5 x 10-3 M

Now, pOH = - log [OH-] = - log (7.5 x 10-3) = -(log 7.5 - 3log10) =- (0.87 - 3) = 2.13

Again , pH + pOH = 14 => pH = 14- pOH = 14 -2.13 = 11.87

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