posted by Naomie on .
WOULD THE VOLUME OF A 0.10M NAOH SOLUTION NEEDED TO TITRATE25.0ML OF A 0.10M HNO2 ( A WEAK ACID)SOLUTION BE DIFFERENT FROMTHAT NEEDED TO TITRATE 25.0ML OF A 0.10M HCL (A STRONGACID)SOLUTION?
that depends on the number of moles of the acids that need to be neutralized by the base and the concentration of the base.
1. HNO2 + NaOH --> NaNO2 + H2O
i.e. 1 mole base react with 1 mole acid
the mole for the HNO2 is cv = 0.1Mx0.025L = 0.0025mol = mole of NaOH
therefore, v(NaOH) = n/c = 0.0025/0.1 = 0.025L (25mL)
2. HCl + NaOH --> NaCl + H2O
so again the 1:1 reaction, and the mole for the HCl is cv = 0.1x0.025L = 0.0025mole = mole of NaOH
v(NaOH) = n/c = 0.0025/0.1 = 25mL
so the volume of NaOH required to neutralize the HNO2 and HCl is 25mL. i.e. same volume.