46 9 mL of 0.749 M hydrobromic acid is added lo 37.8 mL of calcium hydroxide, an

Question

46 9 mL of 0.749 M hydrobromic acid is added lo 37.8 mL of calcium hydroxide, and the resulting solution is found to be acidic. 19.0 mL of 0.547 M barium hydroxide is required to reach neutrality. What is the molarity of the original radium hydroxide solution? M 38.7 mL ot 1.78 M hydrochloric acid is added to 29.9 mL of potassium hydroxide. and (he resulting solution is found to be acidic. 18.0 mL of 1.05 M calcium hydroxide is required to reach neutrality. What is the molarity of the original potassium hydroxide solution? M

Explanation / Answer

2 HBr + Ca(OH)2 = 2 H2O + CaBr2

2 HBr + Ba(OH)2 --> 2 H2O + BaBr2

Concentration of hydrobromic acid = 46.9 x 0.749/1000 = 0.03512 moles

Hence 0.5 equivalent Ca or Ba hydroxide is need = 0.03512 x 0.5 = 0.01756 Moles

Concentration of Barium hydroxide = 19 x 0.547/1000 = 0.01039 Moles

Concentration of Calcium hydroxide = 0.01756-0.01039 = 0.007167 moles

0.007167 x 1000/37.8 = 0.1896 mole calcium hydroxide solution is need

2 HCl + Ca(OH)2 --> 2 H2O + CaCl2

Concentration of HCl = 38.7 x 1.78/1000 = 0.06888 moles

Concentration of Calcium hydroxide = 18 x 1.05/1000 =0.0189 moles

0.0189 moles of Calcium hydroxide will react with (0.0189 x 2) 0.0378 moles HCl

Hence, rest concentration of HCl = 0.06888-0.0378 = 0.03108 moles

Concentration of Potassium hydroxide = 0.03108 x1000/29.9 = 1.039 Moles of KOH is need

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