Calculate the pH of the resulting solution if 30.0 mL of 0.300 M HCl(aq) is added to
(a) 35.0 mL of 0.300 M NaOH(aq).
(b) 40.0 mL of 0.350 M NaOH(aq).
Wait so, you are getting 0.65 from the total mL but shouldnt this be 65/1000 or 0.065?
Why did you divide by .65?
To calculate the pH of the resulting solution when an acid and a base are mixed, you need to know the initial concentrations of the acid and base, as well as the volumes of each solution that are mixed.
(a) For the first scenario, you have 30.0 mL of 0.300 M HCl(aq) and 35.0 mL of 0.300 M NaOH(aq). Here's how you can calculate the resulting pH:
1. Find the limiting reactant: To determine which reactant limits the reaction, you need to consider the stoichiometry of the reaction. The balanced equation for the reaction between HCl and NaOH is:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Since the stoichiometric ratio between HCl and NaOH is 1:1, the limiting reactant will be the one with the smallest initial volume. In this case, both solutions have the same volume, so the limiting reactant is HCl.
2. Determine the moles of the limiting reactant: To calculate the moles of HCl, you need to use the equation:
moles = concentration (in M) × volume (in L)
For HCl:
moles of HCl = 0.300 M × 0.030 L = 0.009 moles
3. Determine the excess reactant: Since NaOH is in excess, there will be some leftover after the reaction. To find the moles of excess NaOH, use the same equation as above:
moles of NaOH = 0.300 M × 0.035 L = 0.0105 moles
4. Convert the moles of excess NaOH to its final concentration: Now that you know the moles of excess NaOH, you can calculate its final concentration in the resulting solution. The total volume of the resulting solution is the sum of the initial volumes of HCl and NaOH (30.0 mL + 35.0 mL = 65.0 mL = 0.065 L):
final concentration of NaOH = moles of NaOH / total volume
final concentration of NaOH = 0.0105 moles / 0.065 L = 0.161 M
5. Calculate the pOH of the resulting solution: The pOH can be found using the equation:
pOH = -log[OH-]
Since NaOH is a strong base and completely dissociates in water, the concentration of OH- ions is the same as the concentration of NaOH:
[OH-] = 0.161 M
pOH = -log(0.161)
6. Calculate the pH of the resulting solution: The pH can be obtained by subtracting the pOH from 14 (pH + pOH = 14):
pH = 14 - pOH
Therefore, you can calculate the pH of the resulting solution by subtracting the pOH calculated in step 5 from 14.
(b) For the second scenario, you have 40.0 mL of 0.350 M NaOH(aq). Following the same steps as in scenario (a), you can calculate the pH of the resulting solution. Since there is no acid present in this scenario, the pH will be determined solely by the concentration of the base NaOH.
I hope this explanation helps you understand the process of calculating the pH of a resulting solution when acid and base solutions are mixed!
I will do one.
first: moles HCL=.030*.3=.009 moles
moles base=.035*.3=.015 moles
so you have excess of .006 moles OH
pOH= -log (.006/.65)
then pH= 14-pOH