What is the pH of the solution created by combining 12.30 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)?
mL NaOH pH wHCl pH wHC2H3O2
12.30
What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water (like it will be in the experiment you perform in lab)?
mL NaOH pH wHCl pH wHC2H3O2
12.30
NaOH + HCl ==>NaCl + HOH
Strong base + strong acid.
calculate mols NaOH. M x L = ??
calculate mols HCl. M x L = ??
Determine mols NaCl and HOH produced.
Determine excess NaOH OR HCl and pH from that.
For NaOH + HC2H3O2 ==> NaC2H3O2 + HOH
Strong base + weak acid.
Determine mols NaOH. M x L = ??
Determine mols HC2H3O2. M x L = ??
Determine mols NaC2H3O2 formed.
Determine mols HC2H3O2 in excess.
This gives you a weak acid and its salt which is a buffer solution. Determine pH from Henderson-Hasselbalch equation which is pH = pKa + log (base)/(acid).
Post your work if you get stuck.
mols NaOH = 0.1 mol/L x 0.0123 L = 0.00123 mols
mols HCL = 0.1 mol/L x 0.008 L = 0.0008
how do you determine mols NaCl and HOH produced?
To find the pH of a solution, we need to consider the concentrations of the acid and base, as well as the volume of the solution. The pH of a solution can be determined using the equation:
pH = -log[H+]
where [H+] represents the concentration of hydrogen ions in the solution.
Let's start by calculating the concentration of each solution after they are mixed together.
For the first scenario, where 12.30 mL of 0.10 M NaOH(aq) is combined with 8.00 mL of 0.10 M HCl(aq):
To calculate the concentration after mixing, we use the formula:
moles of solute / volume of solution
For NaOH:
moles of NaOH = concentration of NaOH x volume of NaOH
= 0.10 mol/L x 0.0123 L
= 0.00123 mol
The volume of the final solution is the sum of the volumes of NaOH and HCl:
volume of solution = volume of NaOH + volume of HCl
= 0.0123 L + 0.008 L
= 0.0203 L
Now we can calculate the concentration of the NaOH solution after mixing:
Concentration of NaOH = moles of NaOH / volume of solution
= 0.00123 mol / 0.0203 L
= 0.0606 M
Similarly, we can calculate the concentration of HCl after mixing:
moles of HCl = concentration of HCl x volume of HCl
= 0.10 mol/L x 0.008 L
= 0.0008 mol
Concentration of HCl = moles of HCl / volume of solution
= 0.0008 mol / 0.0203 L
= 0.0394 M
Now we have the concentrations of both the base (NaOH) and the acid (HCl) in the mixed solution. To find the pH, we need to calculate the concentration of hydrogen ions ([H+]).
In this case, since NaOH is a strong base, it fully dissociates in water into Na+ and OH- ions. The concentration of OH- ions is equal to the concentration of NaOH, which is 0.0606 M.
HCl is a strong acid, so it also fully dissociates in water into H+ and Cl- ions. The concentration of H+ ions is equal to the concentration of HCl, which is 0.0394 M.
For the scenario of 8.00 mL of 0.10 M HC2H3O2(aq) diluted with 100 mL of water:
The concentration of the acid after dilution can be calculated using the formula:
Concentration after dilution = initial concentration x initial volume / final volume
Concentration of HC2H3O2 after dilution = (0.10 mol/L x 0.008 L) / (0.008 L + 0.100 L)
= (0.0008 mol) / 0.108 L
= 0.0074 M
The concentration of hydrogen ions ([H+]) can be determined by considering the dissociation of HC2H3O2, which is a weak acid. HC2H3O2 partially dissociates in water to form H+ and C2H3O2- ions.
Lastly, to find the pH, we can take the negative logarithm of the hydrogen ion concentration using the equation:
pH = -log[H+]
Substituting the values calculated above, we can now determine the pH for both scenarios.