Why does the addition of 50ml of 0.5M NaOH to 50mL of 0.6M KOH cause the pH to decrease (aka more acidic)?
Because you have reduced the (OH^-) of the 0.6 M KOH by adding an equal volume of 0.5 M NaOH
50 mL x 0.6 M KOH = 30 millimols. pH = 13.78
Adding 50 mL x 0.5 M NaOH = 25 millimols NaOH added
30 millimols + 25 millimols = 55 millimols
55 millimols/100 mL = 0.55 M. pH = 13.74.
Actually, diluting with an equal volume of a less basic solution of NaOH effectively averages the 0.6M and 0.5M to 0.55 molar and 0.55 M is more acidic than 0.6 M (but not by much).
To understand why the addition of 50mL of 0.5M NaOH to 50mL of 0.6M KOH causes the pH to decrease, let's break down the process and calculate the resulting concentration and pH.
First, we need to determine the number of moles of KOH and NaOH being added. We can do this by multiplying the volume (50 mL) by the concentration (0.6 M for KOH and 0.5 M for NaOH).
For KOH:
50 mL x 0.6 M KOH = 30 millimols of KOH
For NaOH:
50 mL x 0.5 M NaOH = 25 millimols of NaOH
Next, we combine the moles of KOH and NaOH added to find the total number of moles in the solution:
30 millimols (KOH) + 25 millimols (NaOH) = 55 millimols total
To find the final concentration, we divide the total number of moles by the final volume of the solution, which is 100 mL since the volumes of KOH and NaOH are equal:
55 millimols / 100 mL = 0.55 M
Now, let's calculate the pH of the solution based on its concentration.
We can start by using the fact that pH = -log10([H+]), where [H+] represents the concentration of hydrogen ions.
In this case, we need to calculate the concentration of hydroxide ions (OH-) since we are dealing with a basic solution.
To find the concentration of OH-, we can use the equation [OH-] = Kw / [H+], where Kw is the water dissociation constant (1 x 10^-14 M^2 at 25°C).
Since we have a basic solution and need to find the concentration of OH-, we can rewrite this equation as [OH-] = Kw / [OH-], since [H+] = [OH-] in a neutral solution.
Simplifying, we have [OH-]^2 = Kw, and taking the square root of both sides, we get [OH-] = sqrt(Kw), where sqrt represents the square root.
Substituting the value of Kw, we have [OH-] = sqrt(1 x 10^-14 M^2), which is approximately 1 x 10^-7 M.
Now, since the solution is more basic with higher OH- concentration, we can calculate the pOH by using pOH = -log10[OH-]:
pOH = -log10(1 x 10^-7) = 7
Finally, we can calculate the pH by subtracting the pOH from 14, since pH + pOH = 14:
pH = 14 - 7 = 7
Therefore, the pH of the solution after adding 50mL of 0.5M NaOH to 50mL of 0.6M KOH is approximately 7, indicating that the solution becomes more acidic compared to the initial pH of 13.78 for 0.6M KOH.