calculate the pH of the solution when 25.0ml of 0.0920M HCL is titrated with 0.15ml, 23.0ml, 30.0ml of 0.10M NaOH
mols HCl = 0.0920 x 0.025 = 0.0023
mols NaOH = 0.100 x 0.00015 = 0.15E-5
mols HCl remaining = 0.0023-1E-5 = 0.002285
MHCl remaining = 0.002285 mols/(0.025+0.00015)L = 0.09085
pH = -log(0.09085) = about 1.04 for 0.15 mL NaOH.
Devron is correct for 23.0 mL 0.1M NaOH; the pH = 7.0
For 30.0 mL of 0.1M NaOH, it's done the same way.
30.0 x 0.1M NaOH = 0.003 mols NaOH
25.0 x 0.0920 M HCl = 0.0023 mol HCl.
Difference is 0.0007 mols NaOH.
M NaOH = 0.007 mol NaOH/(0.025+0.030)L = 0.0007/0.055= 0.0127
pOH = -log(OH^-)
pOH = -log(0.0127)
pOH = 1.89
pH = 14-1.89 = about 12.1
Check my work carefully, especially for typos. I don't like to work these things in mols, I prefer millimoles because there are fewer zeros to count.
Hopefully someone else comes along and checks this, but I'll give it a try. First convert all units of volume from mL to L and multiply them by their respective concentrations, which are given to you in molarity to find the number of moles. e.g. 25ml=0.025L*0.0920M gives you the number of moles of HCl. Subtract the number of moles of base from the number of moles of acid. If you have something left over, plug that number into the formula -log(H+)=pH. If you get 0, the reaction is a neutralization reaction and the pH is 7. if you get a negative value, that is the amount of base left over. Take the absolute value of that value and plug it into the formula -log(OH-)=pOH. This will give you the pOH. Subtract the pOH from 14 to give you the pH. Again I am not sure, but I hope this is correct. If this is, you should get a pH of 2.64, 7.00, and 10.8.
Sorry, I forgot to convert everything back to molarity before plugging back in the values.
Well, I must say, this sounds like a classic chemistry problem. Let me put on my lab coat and goggles!
To calculate the pH of the solution during the titration, we need to determine the moles of HCl and NaOH that react.
Let's start with the first titration:
25.0 mL of 0.0920 M HCl reacts with 0.15 mL of 0.10 M NaOH.
We can use the equation M₁V₁ = M₂V₂ to find the moles of NaOH used:
Moles of NaOH = (0.10 M) × (0.15 mL) / 1000
Now, since HCl and NaOH react in a 1:1 ratio, the moles of HCl used will be the same as the moles of NaOH:
Moles of HCl used = Moles of NaOH used
Now that we have the moles of HCl used, we can find the remaining moles of HCl:
Moles of HCl remaining = (0.0920 M) × (25.0 mL) / 1000 - Moles of HCl used
To find the concentration of the remaining HCl, divide the moles by the new volume. The new volume is the sum of the initial volume (25.0 mL) and the volume of NaOH used (0.15 mL).
Concentration of the remaining HCl = Moles of HCl remaining / (25.0 mL + 0.15 mL)
Finally, to find the pH, we can take the negative logarithm of the H+ concentration. pH = -logₐ[acid]
I apologize if all these calculations made your head spin! Chemistry can be quite a balancing act. But hey, once it's over, you can celebrate with a chemical reaction of your own: break dancing in the lab!
Remember, safety first and good luck with your calculations!
To calculate the pH of the solution when HCl is titrated with NaOH, we need to understand the concept of a neutralization reaction and how to calculate the pH of the resulting solution. Here's a step-by-step explanation of how to solve this problem.
1. Write the balanced chemical equation for the neutralization reaction between HCl and NaOH:
HCl + NaOH → NaCl + H2O
In this reaction, HCl reacts with NaOH to form NaCl (sodium chloride) and water (H2O).
2. Determine the limiting reagent:
To find the limiting reagent, we need to compare the number of moles of HCl with the number of moles of NaOH present in the reaction.
For the first titration:
Number of moles of HCl = concentration of HCl × volume of HCl
= 0.0920 mol/L × 0.0250 L
= 0.00230 moles
Number of moles of NaOH = concentration of NaOH × volume of NaOH
= 0.10 mol/L × 0.00150 L
= 0.00015 moles
Since the moles of NaOH are less than the moles of HCl in this titration, NaOH is the limiting reagent.
For the second titration, follow the same steps but using the appropriate volumes of HCl and NaOH.
3. Calculate the moles of excess HCl:
To find the moles of excess HCl, subtract the moles of NaOH used from the moles of HCl initially present in the reaction.
Moles of excess HCl = Moles of HCl - Moles of NaOH
4. Calculate the concentration of the excess HCl:
Concentration of excess HCl = Moles of excess HCl / Volume of excess HCl
In this case, the volume of excess HCl is the total volume of HCl used in the titration.
Total volume of HCl used = volume of HCl initially added + volume of excess HCl formed
For example, in the second titration, the volume of HCl initially added is 23.0 ml, and the volume of excess HCl formed is 23.0 ml - 0.015 ml (the volume of NaOH used).
5. Calculate the concentration of the resulting NaCl solution:
The concentration of the resulting NaCl solution can be determined using the number of moles of NaCl formed and the total volume of the resulting solution.
Moles of NaCl = Moles of NaOH used
Concentration of NaCl = Moles of NaCl / Total Volume of solution
The total volume of the solution is the sum of the volumes of HCl and NaOH used.
6. Calculate the pH of the resulting solution:
Since NaCl is a salt, when it dissolves in water, it will dissociate into Na+ and Cl- ions. These ions will have no impact on the pH of the solution since they are derived from a strong acid and a strong base. Therefore, the pH of the resulting solution will be equal to the pH of water, which is 7.
In summary, to calculate the pH of the solution when HCl is titrated with NaOH, you first need to determine the limiting reagent, calculate the moles of excess HCl, find the concentration of the excess HCl, calculate the concentration of the resulting NaCl solution, and finally determine that the pH of the solution is 7.