What is the pH of the solution created by combining 0.80 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)?
I thought it should be 7? It's not the right answer though.
You have two questions here.
#1. pH of 0.8 mL x 0.1 M NaOH + 8.00 mL of 0.1 M HCl.
moles NaOH = M x L = 0.1 x 0.0008 = 0.00008 moles NaOH.
moles HCl = M x L = 0.1 x 0.008 = 0.0008 moles
NaOH + HCl ==> NaCl + H2O
Now place the moles under the reactants so you can see what is going on. You should note that HCl is in excess; therefore, moles NaCl and H2O formed are 0.00008 and moles HCl left unreacted is 0.0008 - 0.00008 = 0.00072 moles HCl in a total volume of 8.00 + 0.8 mL = 0.0088 L. .
M = moles/L = 0.00072/0.0088 = 0.0818 M
Then pH = -log(H^+) = ??
For the second one, you have HC2H3O2 (a weak acid) and NaOH and the HC2H3O2 is in excess which means that you will have at the end of the reaction a buffered solution consisting of NaC2H3O2 and HC2H3O2. You must use the Henderson-Hasselbalch equation to solve for the pH.
Post your work if you get stuck.
so the henderson eqn is:
pH= pKa +log([conj. base]/[acid])
so do i have to look up pKa for NaOH and then how do i find the conjugate base concentration?
To find the pH of the solution created by combining different solutions, we need to consider the concept of neutralization and the properties of acids and bases.
In this case, we have two scenarios:
Scenario 1: Combining NaOH(aq) with HCl(aq)
Step 1: Calculate the moles of NaOH and HCl using the formula: moles = concentration * volume (in L).
- For NaOH:
moles of NaOH = 0.10 M * (0.80 mL / 1000 mL/ L) = 0.080 mmol
- For HCl:
moles of HCl = 0.10 M * (8.00 mL / 1000 mL/ L) = 0.800 mmol
Step 2: Determine the limiting reagent. The limiting reagent is the one that is completely consumed and controls the reaction. In this case, NaOH is the limiting reagent since it has fewer moles than HCl.
Step 3: Use the balanced chemical equation to determine the reaction stoichiometry. The balanced equation for the reaction between NaOH and HCl is:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
According to the stoichiometry, 1 mole of NaOH reacts with 1 mole of HCl to form 1 mole of NaCl and 1 mole of water.
Step 4: Calculate the moles of any remaining excess reactant after the limiting reagent is consumed. In this case, there is no excess reactant since NaOH is the limiting reagent.
Step 5: Calculate the concentration of H+ ions in the solution. Since NaOH is a strong base and completely dissociates in water to form Na+ and OH- ions, we can conclude that OH- ions react with H+ ions from HCl in a 1:1 ratio.
Therefore, the concentration of H+ ions is equal to the concentration of OH- ions produced by NaOH:
[H+] = [OH-] = moles of NaOH / total volume of the solution (in L)
[H+] = [OH-] = 0.080 mmol / (0.80 mL + 8.00 mL) / 1000 mL/ L = 0.665 mM
To convert the concentration from millimolar (mM) to the pH scale, we use the formula:
pH = -log[H+]
pH = -log(0.665 * 10^-3) ≈ 3.18
Therefore, the pH of the solution created by combining 0.80 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq) is approximately 3.18.
Scenario 2: Combining NaOH(aq) with HC2H3O2(aq)
The process to determine the pH of this solution is similar to Scenario 1. Repeat Steps 1-5 using the appropriate concentrations and volumes for HC2H3O2.