A. An aqueous solution of potassium hydroxide is standardized by titration with

Question

A. An aqueous solution of potassium hydroxide is standardized by titration with a .116 M solution of hydrobromic acid. if 11.4 mL of bees are required to neutralize 11.3 mL of the acid what is the molarity of the potassium hydroxide solution?
B. An aqueous solution of nitric acid is standardized by titration with a .181 M solution of calcium hydroxide. If 21.2 mL of base are required to neutralize 19.1 mL of the acid what is the molarity of the nitric acid solution?
C. An aqueous solution of barium hydroxide is standardized by titration with a .159 M solution of nitric acid. If 24.6 mL of base are required to neutralize 10.5 mL of the acid what is the molarity of the barium hydroxide solution? A. An aqueous solution of potassium hydroxide is standardized by titration with a .116 M solution of hydrobromic acid. if 11.4 mL of bees are required to neutralize 11.3 mL of the acid what is the molarity of the potassium hydroxide solution?
B. An aqueous solution of nitric acid is standardized by titration with a .181 M solution of calcium hydroxide. If 21.2 mL of base are required to neutralize 19.1 mL of the acid what is the molarity of the nitric acid solution?
C. An aqueous solution of barium hydroxide is standardized by titration with a .159 M solution of nitric acid. If 24.6 mL of base are required to neutralize 10.5 mL of the acid what is the molarity of the barium hydroxide solution?
B. An aqueous solution of nitric acid is standardized by titration with a .181 M solution of calcium hydroxide. If 21.2 mL of base are required to neutralize 19.1 mL of the acid what is the molarity of the nitric acid solution?
C. An aqueous solution of barium hydroxide is standardized by titration with a .159 M solution of nitric acid. If 24.6 mL of base are required to neutralize 10.5 mL of the acid what is the molarity of the barium hydroxide solution?

Explanation / Answer

We know number of moles= molarity * volume

A.KOH+ HBr---> KBr+ H2O

Number of moles of HBr= number of moles of KOH

0.116 mol* 11.3 mL = M * 11.4 mL

===> M = 0.115 M

B. 2HNO3+Ca(OH)2----> Ca(NO3)2 + 2H2O

2*number of moles of HNO3= number of moles of Ca(OH)2

2*M*19.1 mL = 0.159 M *24.6 mL

M= 0.102 M

C. 2HNO3+ Ba(OH)2 ----> Ba(NO3)2+. 2H2O

2 moles of HNO3 =1 mole of Ba(OH)2

2*0.159M*10.5 mL = M * 24.6 mL

M= 0.136 M

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