How many milliliters of 8.40×10^2M Ba(OH)2(aq) are required to titrate 48.65mL of 5.26×10^2M HNO3?
volume acid*normality acid=volumebase*Normally base
48.65ml*5.26E2M=2*V*8.40E2M
solve for Volume, V
48.65ml*5.26E2M=2*V*8.40E2M
solve for Volume, V