In a titration experiment a student uses 1.4 m hbr solution and the indicator phenolphthalein to determine the concentration of
a koh solution.
2 answers:
The question is incomplete. Complete question is attached below:
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Answer:
Given: conc. of HBr = 1.4 M
Volume of HBr = 15.4 mL
Volume of KOH = 22.10 mL
We know that, M1V1 = M2V2
(HBr) (KOH)
Therefore, M2 = M1V1/V2
= 1.4 X 15.4/22.10
= 0.9756 M
Concentration of KOH is 0.9756 M.
...........................................................................................................................
Answer:
Given: conc. of HBr = 1.4 M
Volume of HBr = 15.4 mL
Volume of KOH = 22.10 mL
We know that, M1V1 = M2V2
(HBr) (KOH)
Therefore, M2 = M1V1/V2
= 1.4 X 15.4/22.10
= 0.9756 M
Concentration of KOH is 0.9756 M.
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Answer : The correct option is, (b) 22.5 M
Explanation : Given,
Mass of calcium nitrate tetrahydrate = 266 g
Molar mass of calcium nitrate tetrahydrate = 236.15 g/mole
Volume of solution =
Molarity : It is defined as the moles of solute present in one liter of solution.
Formula used :
Now put all the given values in this formula, we get:
As calcium nitrate tetrahydrate dissociate to give 1 mole of calcium ion, 2 moles of nitrate ion and 4 moles of water.
The concentration of nitrate ion =
Thus, the concentration of nitrate ion is, 22.5 M