Calculate the pH reaches 30.0 mL one 0.200 M solution of HCl salt of amitriptyline is added 100.0 mL of a NaOH solution of concentration 80 mM.
To calculate the pH of the resulting solution, we need to determine the concentration of the resulting substances.
First, let's calculate the number of moles of HCl and NaOH added.
For HCl:
Moles of HCl = Molarity × Volume
Moles of HCl = 0.200 M × 0.0300 L
Moles of HCl = 0.006 mol
For NaOH:
Moles of NaOH = Molarity × Volume
Moles of NaOH = 0.080 M × 0.100 L
Moles of NaOH = 0.008 mol
Next, we need to determine the limiting reactant, which is the one that will be completely consumed in the reaction. In this case, HCl and NaOH react in a 1:1 ratio according to the balanced equation:
HCl + NaOH → NaCl + H2O
Since both HCl and NaOH are in a 1:1 ratio, the limiting reactant will be the one with the smaller number of moles, which is 0.006 mol of HCl.
Now, let's calculate the moles of excess NaOH remaining after the reaction:
Moles of excess NaOH = Moles of NaOH added - Moles of HCl used
Moles of excess NaOH = 0.008 mol - 0.006 mol
Moles of excess NaOH = 0.002 mol
To find the final concentration of the excess NaOH, we need to determine the total volume of the resulting solution:
Total volume = Volume of HCl + Volume of NaOH
Total volume = 0.030 L + 0.100 L
Total volume = 0.130 L
Concentration of excess NaOH (in Molarity):
Concentration = Moles of excess NaOH / Total volume
Concentration = 0.002 mol / 0.130 L
Concentration = 0.0154 M
Now, to find the pH, we need to calculate the pOH using the concentration of the excess NaOH:
pOH = -log[OH-]
pOH = -log(0.0154)
pOH ≈ 1.81
Finally, to find the pH, we subtract the pOH from 14:
pH = 14 - pOH
pH = 14 - 1.81
pH ≈ 12.19
Therefore, the pH of the resulting solution is approximately 12.19.