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

Consider two solutions. One solution is 0.1385 M Ba(OH)2. The other 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 HBr solution, what is the pH of the resulting solution?

Expert Solution

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))

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

= 0.009687 so resulting solution is basic hence

pOH = -log(0.009687)

pOH = 2.0138

pH = 14 - 2.0138 = 11.986

queen_honey_blossom answered 2 years ago

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