Asked by Naomie
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?
Answers
Answered by
bonjo
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.
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.