Question

1) Calculate the pH of the resulting solution if 35.0 mL of 0.350 M HCl(aq) is added to 40.0 mL of 0.350 M NaOH(aq) .

2) Calculate the pH of the resulting solution if 35.0 mL of 0.350 M HCl(aq) is added to 45.0 mL of 0.400 M NaOH(aq) .

Answer 1

2)HCl is a strong acid so it completely dissolves into H+ and Cl- so let's see how much moles of H+ there are: .350 M x .0350 = .012 moles of HCl and thus H+

Next, the same goes for NaOh, just its a base, .40 M x .045 L = .018 moles NaOh and thus of OH-

So .012 out of the .018 moles of OH will react with all the .012 H+ and there will be left .006 OH.

Now we find the molarity of OH: .006/( .035 + .045) L = .075 M

Now we find pH, - Log [H+], or 14 minus -log [OH]: -log .075 = 1.124 ----> 14 - 1 = 12.87~13
13 = pH.

1)HCl is a strong acid so it completely dissolves into H+ and Cl- so let's see how much moles of H+ there are: .350 M x .0350 = .012 moles of HCl and thus H+

Next, the same goes for NaOh, just its a base, 0.35 M x .040 L = .014 moles NaOh and thus of OH-

So .012 out of the .014 moles of OH will react with all the .012 H+ and there will be left .002 OH.

Now we find the molarity of OH: 0.002/( .035 + .040) L = .026 M

Now we find pH, - Log [H+], or 14 minus -log [OH]: -log .026 = 1.58 ----> 14 - 1.58 = 12.41
13 = pH.