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A 0.5 g sample of strontium chloride required 45.6 ml of silver nitrate. silver nitrate was prepared by dissolving 5 g pure silver to make 500 ml of solution. excess silver nitrate consumed 16 ml of 0.015 ml of potassium thiocyanate. calculate the normality of the silver nitrate solution
Accepted Answer
The normality of the silver nitrate solution can be calculated using the following steps:
1. Calculate the moles of silver nitrate used in the titration:
Moles of AgNO3 = (45.6 mL AgNO3) x (1 L / 1000 mL) x (0.015 mol AgNO3 / 1 L)
= 0.000684 mol AgNO3
2. Calculate the moles of silver nitrate consumed by the excess silver nitrate:
Moles of AgNO3 consumed by excess AgNO3 = (16 mL KSCN) x (1 L / 1000 mL) x (0.015 mol KSCN / 1 L)
= 0.00024 mol AgNO3
3. Calculate the total moles of silver nitrate used:
Total moles of AgNO3 = moles of AgNO3 used in the titration + moles of AgNO3 consumed by the excess AgNO3
= 0.000684 mol AgNO3 + 0.00024 mol AgNO3
= 0.000924 mol AgNO3
4. Calculate the normality of the silver nitrate solution:
Normality of AgNO3 = (Total moles of AgNO3) / Volume of AgNO3 solution (in liters)
= 0.000924 mol AgNO3 / 0.5 L
= 0.001848 N
Therefore, the normality of the silver nitrate solution is 0.001848 N.
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