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In A Titration Experiment 50.0 Ml Of 0.375 M Hi Solution Is Required To Neutralize 35.0 Ml Of Ca Oh

How many milliliters of 0.165 M HCl are needed to neutralize completely 35.0 mL of 0.101 M Ba(OH)2 solution?

Moles Ba(OH)2 = 0.101 M x 0.0350 = 0.003535
Moles OH- = 2 x 0.003535 =0.007070
= moles HCl needed
V = 0.007070 / 0.165 M = 0.0428 L=> 42.8 mL

In a titration experiment 50.0 mL of 0.375 M HI solution is required to neutralize 35.0 mL of Ca(OH)2 solution. What is the concentration of?

In a titration experiment 50.0 mL of 0.375 M HI solution is required to neutralize 35.0 mL of Ca(OH)2 solution. What is the concentration of the Ca(OH)2 solution?

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