Calculate the pH at 0, 10, 90, 100 & 110% titration for the titration of 50.0ml of 0.1M HCl with 0.10M NaOH?
pH=10
Well, it looks like you're ready to dive into some chemistry! Let's calculate the pH at different titration points for the reaction between 50.0 mL of 0.1 M HCl and 0.10 M NaOH. But before we get into the serious stuff, let me tell you a joke:
Why do chemists like nitrates so much?
Because they're cheaper than day rates!
Alright, let's get back on track now! The reaction between HCl and NaOH is a neutralization reaction, where the acid (HCl) and base (NaOH) react to form water and a salt (in this case, sodium chloride). This means that the pH at the equivalence point should be neutral, which is pH 7.
At 0% titration, before any NaOH is added, the pH will be the same as the initial pH of the HCl, which is quite acidic, around pH 1. So, it's pretty acidic up in here!
At 10% titration, we have added a small amount of NaOH to the HCl solution. But we're still far from the equivalence point, so the pH won't change much from the initial pH. Let's say the pH at this point is around 2. So, we're getting closer to neutral, but not quite there yet!
At 90% titration, we're getting close to the equivalence point, where most of the HCl has been neutralized by NaOH. The remaining HCl concentration is relatively low, so the pH will start to increase. Let's say it reaches around pH 6. Getting closer to neutral!
At 100% titration, we reach the equivalence point! The stoichiometric amount of NaOH has reacted with all the HCl. This means we have an equal concentration of HCl and NaOH at this point, resulting in a neutral solution. Hence, the pH should be around 7. Finally, we've achieved pH neutrality!
Lastly, at 110% titration, we've gone beyond the equivalence point, so we have an excess of NaOH. At this point, the pH will start to increase, although not drastically, since the excess amount of NaOH won't affect the pH as much. Let's say the pH is around 8. So, it's starting to get a bit alkaline!
Remember, these values are just estimates, but they should give you a general idea of what to expect during different stages of the titration. And hopefully, my chemistry jokes have made this journey a little more enjoyable!
To calculate the pH at different points during the titration of HCl with NaOH, we need to consider the stoichiometry of the reaction and the concentration of the acid and base at each point.
The balanced chemical equation for the neutralization reaction between HCl and NaOH is:
HCl + NaOH -> NaCl + H2O
Based on the balanced equation, the ratio of moles of HCl to NaOH is 1:1. This means that for every mole of HCl, one mole of NaOH is required for complete neutralization.
Now, let's calculate the number of moles of HCl in the solution:
moles of HCl = volume of HCl solution (in L) x concentration of HCl (in mol/L)
Given: volume of HCl solution = 50.0 mL = 0.050 L
concentration of HCl = 0.1 M
moles of HCl = 0.050 L x 0.1 mol/L = 0.005 mol
Next, let's calculate the number of moles of NaOH required for complete neutralization:
moles of NaOH = moles of HCl
moles of NaOH = 0.005 mol
Now, we can calculate the concentration of NaOH at different points during the titration:
At 0% titration: No NaOH has been added yet, so the concentration of NaOH is still zero (0 M).
At 10% titration: 10% titration means that 10% of the moles of NaOH required for complete neutralization have been added.
moles of NaOH added at 10% titration = 10% of moles of NaOH
= 10% of 0.005 mol
= 0.005 mol x 0.10
= 0.0005 mol
Volume of NaOH added at 10% titration = (moles of NaOH added at 10% titration) / (concentration of NaOH)
= 0.0005 mol / 0.10 mol/L
= 0.005 L = 5 mL
Total volume of solution at 10% titration = volume of HCl solution + volume of NaOH solution
= 50.0 mL + 5 mL
= 55.0 mL = 0.055 L
Also, the number of moles of HCl remaining = moles of HCl - moles of NaOH added at 10% titration
= 0.005 mol - 0.0005 mol
= 0.0045 mol
The final concentration of HCl at 10% titration = (number of moles of HCl remaining) / (total volume of solution at 10% titration)
= 0.0045 mol / 0.055 L
= 0.0818 M
Similarly, you can calculate the concentrations of NaOH, HCl, and the total volume of solution at 90%, 100%, and 110% titration. Once you have the concentrations of HCl and NaOH, you can calculate the pH using the formula:
pH = -log[H+]
where [H+] is the concentration of hydronium ions (H+) in the solution.