Question

To prepare a buffer you weigh out 7.20 grams of NaHCO3 and place it into a 400.00 mL volumetric flask. To this flask you add 56.0 mL of 5.60 M H2CO3 and then fill it about halfway with distilled water, swirling to dissolve the contents. Finally, the flask is filled the rest of the way to the mark with distilled water.What is the pH of the buffer that you have created?Acid KaH2CO3 4.3 X 10⁻⁷ HCN 4.9 X 10⁻¹⁰HNO2 4.6 X 10⁻⁴C6H5COOH 6.5 X 10⁻⁵

Answer 1

pH = 5.80

Step-by-step explanation:

The buffer solution is:

H₂CO₃(aq) + H₂O(l) ⇄ HCO₃⁻Na⁺(aq) + H₃O⁺(aq)

To find the pH of the buffer solution we will use the Henderson-Hasselbalch equation:

`pH = pKa + log(([NaHCO_(3)])/([H_(2)CO_(3)]))` (1)

First, we need to find the concentration of the buffer solution. For the NaHCO₃ we have:

`[NaHCO_(3)] = (mol)/(V) = (m)/(M*V)`

Where:

m: is the mass of the NaHCO₃ = 7.20 g

M: is the molar mass of the NaHCO₃ = 84.007 g/mol

V: is the volume of the solution = 400.0 mL

Hence, the concentration of NaHCO₃ is:

`[NaHCO_(3)] = (7.20 g)/(84.007 g/mol*400.0 \cdot 10^(-3) L) = 0.214 M`

Now, the concentration of H₂CO₃ is:

`V_(i)C_(i) = V_(f)C_(f)`

Where:

Vi: is the initial volume of H₂CO₃ = 56.0 mL

Ci: is the initial concentration of H₂CO₃ = 5.60 M

Vf: is the final volume of H₂CO₃ = 400.0 mL

Cf: is the final concentration of H₂CO₃ (to find)

`C_(f) = (V_(i)C_(i))/(V_(f)) = (56.0 mL*5.60 M)/(400.0 mL) = 0.784 M`

Finally, we can use the equation (1) to find the pH of the buffer solution:

`pH = -log(4.3 \cdot 10^(-7)) + log((0.214 M)/(0.784 M)) = 5.80`

I hope it helps you!