Posted by Kate on Monday, March 26, 2012 at 8:05pm.
How many mL of a .10 M NaOH solution are needed to neutralize 15 mL of .20 M H3PO4 solution?
Write nbalanced equation:
3NaOH + H3PO4 → Na3PO4 + 3H2O
3mol NaOH recat with 1mol H3PO4
In all these titration questions, do it the simple way: Use the following equation:
Ma*Va/na = Mb*Vb/nb
Where:
Ma = molarity of acid
Va = volume of acid
na = number of moles acid used, from the balanced equation
Mb = molarity of base
Vb = volume of base
nb = number of moles of base used, from the balanced equation:
substitute:
0.20*15/1 = 0.10*Vb/3
Vb = 0.20*15*3 /0.10
Vb = 90ml
The correct answer is : volume of 0.10NaOH required = 90ml.
___________________________________
Similar problem, different concentrations, volumes, etc.
Just try it yourself! :)
How many mL of a .10 M NaOH solution are needed to neutralize 15 mL of .20 M H3PO4 solution?
Write nbalanced equation:
3NaOH + H3PO4 → Na3PO4 + 3H2O
3mol NaOH recat with 1mol H3PO4
In all these titration questions, do it the simple way: Use the following equation:
Ma*Va/na = Mb*Vb/nb
Where:
Ma = molarity of acid
Va = volume of acid
na = number of moles acid used, from the balanced equation
Mb = molarity of base
Vb = volume of base
nb = number of moles of base used, from the balanced equation:
substitute:
0.20*15/1 = 0.10*Vb/3
Vb = 0.20*15*3 /0.10
Vb = 90ml
The correct answer is : volume of 0.10NaOH required = 90ml.
___________________________________
Similar problem, different concentrations, volumes, etc.
Just try it yourself! :)
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