75 ml of koh is titrated with 0.19 m hbr. 65 ml of hbr is added to reach the equivalence point what is the concentration of koh?"
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This is a reaction between an alkali and an acid, and the products will be salt and water.
The balanced equation for the reaction is:
KOH (aq) + HBr(aq) --------> KBr (aq) + H2O (l)
From the above equation, the mole ratio KOH : HB = 1: 1
∴ At the equivalence point, No. of moles of KOH = No. of moles of HBr
No. of moles of HBr = volume in litres x molarity = 65/1000 x 0.19 = 0.01235
Moles of KOH = 0.01235 = 75/1000 x Molarity of KOH
Molarity of KOH = (0.01235 x 1000)/75
= 0.165 M
∴ the concentration of KOH = 0.165 M
The balanced equation for the reaction is:
KOH (aq) + HBr(aq) --------> KBr (aq) + H2O (l)
From the above equation, the mole ratio KOH : HB = 1: 1
∴ At the equivalence point, No. of moles of KOH = No. of moles of HBr
No. of moles of HBr = volume in litres x molarity = 65/1000 x 0.19 = 0.01235
Moles of KOH = 0.01235 = 75/1000 x Molarity of KOH
Molarity of KOH = (0.01235 x 1000)/75
= 0.165 M
∴ the concentration of KOH = 0.165 M
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