What is the pH value if you take into account that the 8.00 mL of 0.10 M HCl was first diluted with 100 mL of water
and titrated with 1.60 mL of the 0.10 M NaOH
i know that the concentration of HCl after diluting is .008M but than im not sure how to convert that to mols which volume do i use
Won't both give you the same answer. After all, the number of mols is not changing. SO M x L = mols.
0.1 M x 0.008 L = ??
0.008 M x 0.1 L (that's 100 mL) = ??
The (HCl) at the for this problem, after the reaction, simply is moles/L but the liters is now 100 mL + 1.60 mL + 8.00 (check these numbers since I'm not looking at them--that's just the way I remember the problem).
No, that is not the new dilution. The new volume is 108ml (you added 100ml) with the .1*8/1000 moles of HCL.
Now, the NaOH will remove (.0016*.1) moles in titration.
New moles of HCL= .1*.008-.0016*.1
volume= 108+1.6 ml
New concentration of H+ is molesHCl/volume in liters.
To calculate the pH value after dilution and titration, you first need to determine the mols of HCl in the diluted solution.
1. Start by calculating the number of mols of HCl in the initial 8.00 mL of 0.10 M HCl solution:
Mols of HCl = concentration (in M) x volume (in L)
Mols of HCl = 0.10 M x 0.008 L = 0.0008 mol
2. Next, calculate the volume of the diluted solution:
Volume of diluted solution = initial volume + diluent volume
Volume of diluted solution = 8.00 mL + 100 mL = 108.00 mL = 0.108 L
3. Now, determine the concentration of HCl in the diluted solution:
Concentration of HCl in the diluted solution = mols of HCl / volume of diluted solution
Concentration of HCl in the diluted solution = 0.0008 mol / 0.108 L = 0.0074 M
4. Since the volume of NaOH used for titration was 1.60 mL, you need to calculate the number of mols of NaOH used:
Mols of NaOH = concentration (in M) x volume (in L)
Mols of NaOH = 0.10 M x 0.00160 L = 0.00016 mol
5. The balanced equation for the reaction between HCl and NaOH is:
HCl + NaOH → NaCl + H2O
From the equation, you can see that 1 mol of HCl reacts with 1 mol of NaOH.
6. Since we know that 0.00016 mol of NaOH was used, this means that 0.00016 mol of HCl also reacted.
7. To determine the remaining mols of HCl, subtract the mols of HCl that reacted from the initial mols:
Remaining mols of HCl = initial mols - mols of HCl that reacted
Remaining mols of HCl = 0.0008 mol - 0.00016 mol = 0.00064 mol
8. Finally, calculate the concentration of HCl in the titrated solution:
Concentration of HCl in the titrated solution = remaining mols of HCl / volume of diluted solution
Concentration of HCl in the titrated solution = 0.00064 mol / 0.108 L = 0.0059 M
Now that you have the concentration of HCl in the titrated solution, you can use it to calculate the pH value.