Calculate the volume of .100M NaOH that must be added to reach ph of 3 in the titration of 25.00 mL of .100M HCl
mols HCl initially = 0.025 x 0.1 = 0.0025
To end up with pH = 3 (0.001M H^+) we want H^+ to be 1E-3. If we work in millimols and let x = mL of 0.1M NaOH, we have then
[(25.00 x 0.1M)-(0.1x)/(25+x)] = 1E-3M
Solve for x and I obtained approx 24.5 mL of 0.1M NaOH that must be addd. We can check to see if that is right and we have
25.00 x 0.1M HCl = 2.5 mmols initially.
-24.50 x 0.1M NaOH = 2.45mmols NaOH add
-------
0.5 millimols HCl reamins. (HCl) = 0.5 mmol/(24+24.5 mL) = 0.00101
for which pH = 2.996. If you carry the numbers another place or two in the calculations above you will get 1E-3 and pH 3.
To calculate the volume of 0.100M NaOH that must be added to reach a pH of 3 in the titration of 25.00 mL of 0.100M HCl, you need to consider the stoichiometry of the reaction and the equivalence point of the titration.
In this case, the reaction between HCl and NaOH can be represented as follows:
HCl + NaOH -> NaCl + H2O
From the balanced equation, you can see that the stoichiometric ratio between HCl and NaOH is 1:1. This means that for every 1 mole of HCl, you will need 1 mole of NaOH to neutralize it.
First, calculate the number of moles of HCl present in the initial solution:
moles of HCl = (concentration of HCl) * (volume of HCl)
moles of HCl = (0.100M) * (0.02500 L) = 0.0025 moles
Since the stoichiometric ratio between HCl and NaOH is 1:1, you will need 0.0025 moles of NaOH to neutralize the HCl.
Now, we need to calculate the volume of 0.100M NaOH required to reach a pH of 3. At pH 3, the concentration of H+ ions is 10^-3 M.
Using the equation for the reaction of NaOH with H+ ions:
NaOH + H+ -> Na+ + H2O
Concentration of NaOH = Concentration of H+
Therefore,
volume of NaOH = (moles of NaOH) / (concentration of NaOH)
volume of NaOH = (0.0025 moles) / (0.100 M) = 0.025 L = 25 mL
Therefore, you would need to add 25.00 mL of 0.100M NaOH to reach a pH of 3 in the titration of 25.00 mL of 0.100M HCl.