What is the pH of the solution created by combining 1.30 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)?
This is how I would go about solving the problem, but could you tell me where I went wrong?
First I would find the number of mols of OH from NaOH and then use
HCl+OH -> HOH + Cl- to find the mols of HCl and Cl- after neutralization...and then use an ICE chart to find H and henceforth pH. I don't think this is correct because my answer came out negative.
The method looks ok.
NaOH + HCl ==> NaCl + HOH
mols NaOH = L x M = (1.30 mL/1000) x0.10 = 0.00013
mols HCl = L x M = (8/1000) x 0.10 = 0.0008
Reagent in excess is HCl. Therefore, all of the NaOH will be used, 0.00013 mols NaCl and HOH formed. Mols HCl remaining unreacted is 0.0008-0.00013 = 0.00067
Molarity HCl remaining is mols/L (combined volume is 1.13 + 8.9 = 9.13 mL)
Then find pH of the HCl solution. Should be about 1 or so.
Your approach is correct in identifying that you need to use an ICE chart to find the concentration of H+ ions and then calculate the pH. However, there seems to be a mistake in your calculation or application of the ICE chart, resulting in a negative value.
Let's go through the process step-by-step to identify the error:
1. Calculate the moles of NaOH:
Moles of NaOH = volume (in L) × molarity
Moles of NaOH = 1.30 mL × (1 L / 1000 mL) × 0.10 mol/L
Moles of NaOH = 0.0013 mol
2. Calculate the moles of HCl:
Moles of HCl = volume (in L) × molarity
Moles of HCl = 8.00 mL × (1 L / 1000 mL) × 0.10 mol/L
Moles of HCl = 0.008 mol
3. Since the reaction between NaOH and HCl is a 1:1 ratio, the moles of HCl and OH- ions will be equal after neutralization.
4. Calculate the total moles of OH- ions:
Total moles of OH- = moles of NaOH
Total moles of OH- = 0.0013 mol
5. Calculate the concentration of OH- ions after mixing:
Volume of final solution = volume of NaOH + volume of HCl = 1.30 mL + 8.00 mL = 9.30 mL = 0.0093 L
Concentration of OH- = moles of OH- / volume of final solution
Concentration of OH- = 0.0013 mol / 0.0093 L
Concentration of OH- = 0.139 M
6. Since the reaction between HCl and OH- ions forms water, all the moles of OH- are consumed, leaving no remaining OH- ions.
7. Calculate the concentration of H+ ions after neutralization:
Concentration of H+ = 0 M (because all OH- ions react completely)
8. Calculate the pH:
pH = -log[H+]
pH = -log(0)
pH = undefined (or you can say pH = infinity)
The mistake in your calculation might be due to the assumption that H+ concentration is not zero after reaction with OH-. In reality, all the OH- ions react with H+ ions to form water, resulting in no remaining OH- or H+ ions. Therefore, the pH of the solution is undefined or infinite.