REVISION TO PREVIOUS POST:
What is the pH of the solution created by combining 1.90 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 understood how to do the HC2H3O2, but I can't do the HCl. I am trying to do the same procedure but I am getting a log(-0.00061), which does not exist.Can someone help me please?
0.0019L mols of NaOH
0.008L x 0.1M HCL = 0.0008 mols HCL
0.00019 - 0.0008 = -0.00061 mols OH left over
Total Volume: 0.0019 + 0.008 = 0.0099
-0.00061 mols/0.0099 L = -0.0616 M [OH-]
pOH = -log [OH-] = -log [-0.0616] = Error
0.0019L mols of NaOH
That was 0.0019 K x 0.1 M = 0.00019 moles NaOH
0.008L x 0.1M HCL = 0.0008 mols HCL
0.00019 - 0.0008 = -0.00061 mols OH left over
The larger number is 0.0008 so it will be left over. 0.0008 HCl - 0.00019 NaOH = +0.00061 moles HCl left over.
Total Volume: 0.0019 + 0.008 = 0.0099
-0.00061 mols/0.0099 L = -0.0616 M [OH-]
pOH = -log [OH-] = -log [-0.0616] = Error
0.00061 moles HCl/0.0099 = 0.0616 M
pH = -log(0.0616) = 1.21
Thank You Dr.Bob. I really appreciate your help.
To find the pH of the solution created by combining NaOH and HCl, we need to understand that this is a neutralization reaction. NaOH is a strong base, and HCl is a strong acid. When they react, they form water (H2O) and a salt (NaCl).
To calculate the pH, we need to determine the concentration of H+ ions in the resulting solution.
Let's start with the reaction between NaOH and HCl:
NaOH(aq) + HCl(aq) ⟶ H2O(l) + NaCl(aq)
Since NaOH and HCl react in a 1:1 ratio, we can conclude that after the reaction, the concentration of H+ ions will be the same as the concentration of OH- ions.
Now, let's calculate the concentration of OH- ions:
First, determine the moles of NaOH:
Moles of NaOH = volume (L) × molarity (mol/L)
= 1.90 mL × 0.10 mol/L
= 0.190 mmol = 0.00019 mol
Since NaOH and HCl are in a 1:1 ratio, the moles of HCl will also be 0.00019 mol.
Next, determine the volume of the resulting solution by combining NaOH and HCl. Add the volumes of NaOH and HCl:
Total volume = 1.90 mL + 8.00 mL
= 9.90 mL = 0.00990 L
Now, calculate the concentration of OH- ions:
OH- concentration (mol/L) = moles / volume (L)
= 0.00019 mol / 0.00990 L
≈ 0.019 mol/L
Since the solution is neutral, the concentration of H+ ions will also be 0.019 mol/L.
Finally, calculate the pH:
pH = -log[H+]
= -log(0.019)
≈ 1.72
Therefore, the pH of the solution created by combining 1.90 mL of 0.10 M NaOH(aq) with 8.00 mL of 0.10 M HCl(aq) is approximately 1.72.