One liter of approximetely 0.25N HNO3 has been prepared. Upon titration it was found that 5.0 ml of the acid required 11.0 ml of 0.10N NaOH for neutrality. How much concetrated HNO3 at 15.8N must be added to one liter to make it exactly 0.25N?
So the first thing is to find the exact concentration of the acid.
concAcid*5=.10*.11
concacid=.22N
So in a liter, you have .22molesAcid
YOu need .25 moles
so you have to add .03 moles acid.
volumeacid=.03*15.8=.474
volume acid=molesacid/molarity=.03/15.8=.190e-3 liters or .190ml acid
check my thinking.
To solve this problem, we need to use the principle of equivalence in acid-base titrations. The principle states that the number of moles of acid is equal to the number of moles of base at the equivalence point.
Let's start by calculating the number of moles of NaOH used in the titration:
Moles of NaOH = volume of NaOH (in liters) x concentration of NaOH
Volume of NaOH = 11.0 ml = 11.0/1000 = 0.011 L
Concentration of NaOH = 0.10 N
Moles of NaOH = 0.011 L x 0.10 N = 0.0011 moles
Since the acid and base react in a 1:1 ratio, the number of moles of HNO3 in 5.0 ml of the acid is also 0.0011 moles.
Now, we can calculate the concentration of the original HNO3 solution before dilution:
Concentration of original HNO3 solution = moles of HNO3 / volume of HNO3 (in liters)
Volume of HNO3 = 5.0 ml = 5.0/1000 = 0.005 L
Concentration of original HNO3 solution = 0.0011 moles / 0.005 L = 0.22 N
We need to add a concentrated HNO3 solution at 15.8 N to make a 1.00 L solution with a concentration of 0.25 N. Let's assume x represents the volume of the concentrated HNO3 solution in liters.
The moles of HNO3 added from the concentrated solution can be calculated as:
Moles of HNO3 added = concentration of HNO3 x volume of HNO3 added
Moles of HNO3 added = 15.8 N x x L
The total moles of HNO3 in the final solution can be calculated as:
Total moles of HNO3 = moles of HNO3 from the original solution + moles of HNO3 added
Total moles of HNO3 = 0.0011 moles + 15.8 N x x L
Since the total volume of the final solution is 1.00 L, we can calculate the concentration of the final solution:
Concentration of final HNO3 solution = Total moles of HNO3 / Volume of final solution
Concentration of final HNO3 solution = (0.0011 moles + 15.8 N x x L) / 1.00 L
We know that the final concentration should be 0.25 N, so we can set up the following equation:
0.25 N = (0.0011 moles + 15.8 N x x L) / 1.00 L
Solving this equation will give us the volume of the concentrated HNO3 solution needed to make the final solution. The resulting volume should be subtracted from 1.00 L to find the volume of water that needs to be added to the final solution to reach one liter.
I recommend using a calculator or a computer program to solve the equation, as it involves algebraic manipulation and calculations.