Consider two solutions. One solution is 0.1385 M Ba(OH)2. The other is 0.2050 M HBr. a. Calculate...

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Chemistry

Consider two solutions. One solution is 0.1385 M Ba(OH)2. Theother is 0.2050 M HBr.
a. Calculate the pH of the HBr solution.
b. Calculte the pH of the Ba(OH)2 solution.
c. If 50.00 mL of the Ba(OH)2 is mixed with 30.00 mL of the HBrsolution, what is the pH of the resulting solution?

Answer & Explanation Solved by verified expert

3.8 Ratings (773 Votes)

pH = -log (0.2050)

pH = 0.688

pOH = -log(0.1385)

pOH = 0.8585

pH = 14 - pOH

pH = 14 - 0.8585

pH = 13.1415

Resulting solution concentration = ((M1V1 - M2V2) / (V1 + V2))

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (((0.1385*50) - (0.205*30)) / (30+50))

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.009687 so resulting solution is basic hence

pOH = -log(0.009687)

pOH = 2.0138

pH = 14 - 2.0138 = 11.986

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