What are the pH of the solutions?
a). 0.10 mol of NaOH is added to 1.0 L of a solution that is 0.40 M CH3NH2 and 0.70 M CH3NH3Cl
b). 0.10 mol of HCl is added to 1.0 L of a solution that is 0.40 M CH3NH2 and 0.70 M CH3NH3Cl
a). 0.10 mol of NaOH is added to 1.0 L of a solution that is 0.40 M CH3NH2 and 0.70 M CH3NH3Cl
b). 0.10 mol of HCl is added to 1.0 L of a solution that is 0.40 M CH3NH2 and 0.70 M CH3NH3Cl
1 Answer
They are both buffers. The starting
"pH" = "pK"_a + log\frac(["B"])(["BH"^(+)])
You have not supplied the
"pK"_a = -log(K_w/K_b) = ???
Adding
color(blue)("pH") = "pK"_a + log(("0.40 M B" cdot "1.0 L" + "0.10 mols NaOH")/("0.70 M BH"^(+) cdot "1.0 L" - "0.10 mols NaOH"))
= color(blue)("pK"_a - 0.079)
You can convince yourself that adding
color(blue)("pH") = "pK"_a + log(("0.40 M B" cdot "1.0 L" - "0.10 mols HCl")/("0.70 M BH"^(+) cdot "1.0 L" + "0.10 mols HCl"))
= color(blue)("pK"_a - 0.426)