math,science,lab
posted by jc .
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?

HNO3+NaOH=NaNO3+H2O
No. of moles of NaOH = 0.011*0.1 = 0.0011M
Therefore no. of moles of HNO3 in 5ml.
= 0.0011M
Concentration of diluted HNO3
= 0.0011/0.005=0.22N
Volume of 15.8N HNO3 required = x litre.
0.22N*1 litre+15.8x = 0.25(1+x)
Solve for x to get
x=3/1555
=0.00193 l.
=1.93 ml.