Question

A 110.0 mL buffer solution is 0.110 molL−1 in NH3 and 0.125 molL−1 in NH4Br.

(A) What mass of HCl will this buffer neutralize before the pH falls below 9.00? (Kb(NH3)=1.8×10−5)?

(B) If the same volume of the buffer were 0.265 molL−1 in NH3 and 0.390 molL−1 in NH4Br, what mass of HCl could be handled before the pH fell below 9.00?

Answer 1

(A)

Molarity of NH₃ = 0.110 M

Volume of the buffer solution = 110 mL

Moles of NH₃ = 0.110 x 110 = 12.1 millimoles = 0.0121 moles

Molarity of NH₄⁺ = 0.110 M

Volume of the buffer solution = 110 mL

Moles of NH₄⁺ = 0.125 x 110 = 13.75 millimoles = 0.01375 moles

Let us say that this buffer neutralizes Y moles of HCl before the pH falls below 9.00.

Then after Y moles of HCl, Y moles of NH₃ is consumed to produce Y moles of NH₄⁺.

Thus, the new moles of NH₃ = (0.0121 - Y) mol and moles of NH₄⁺ = (0.01375 + Y) mol.

Total volume of the buffer solution = 110 mL = 0.110 L     (Assuming that volume change due HCl is negligible)

After the new concentration of NH₃ = (0.0121 - Y)/0.110 M and moles of NH₄⁺ = (0.01375 + Y)/0.110 M

Given, K_b(NH₃)=1.8×10⁻⁵

Now, K_a (NH₄⁺) = K_w/K_b

                          = 10^-14/1.8×10⁻⁵

                          = 5.6 x 10^-10

Therefore, pK_a = - logK_a = - log5.6 x 10^-10 = 9.26

From Henderson-Hasselbalch equation,

pH = pK_a + log[base]/[acid]

or, 9.00 = 9.26 + log {(0.0121 - Y) x 0.110}/{(0.01375 + Y) x 0.110}

or, log (0.0121 - Y) /(0.01375 + Y) = - 0.26

or, (0.0121 - Y) /(0.01375 + Y) = 0.5495

or, (0.0121 - Y) = 0.00756 + 0.5495Y

or, 1.5495Y = 0.00454

or, Y = 0.00293

Thus the moles of HCl = 0.00293 mol

Molar mass of HCl = 36.5 g/mol

Therefore, the mass of HCl = 36.5 x 0.00293 = 0.107 g

Therefore, this buffer neutralizes 0.107 g of HCl before the pH falls below 9.00.

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(B)

Molarity of NH₃ = 0.265 M

Volume of the buffer solution = 110 mL

Moles of NH₃ = 0.265 x 110 = 29.15 millimoles = 0.02915 moles

Molarity of NH₄⁺ = 0.390 M

Volume of the buffer solution = 110 mL

Moles of NH₄⁺ = 0.390 x 110 = 42.9 millimoles = 0.04290 moles

Let us say that this buffer neutralizes Y moles of HCl before the pH falls below 9.00.

Then after Y moles of HCl, Y moles of NH₃ is consumed to produce Y moles of NH₄⁺.

Thus, the new moles of NH₃ = (0.02915 - Y) mol and moles of NH₄⁺ = (0.04290 + Y) mol.

Total volume of the buffer solution = 110 mL = 0.110 L     (Assuming that volume change due HCl is negligible)

After the new concentration of NH₃ = (0.02915 - Y)/0.110 M and moles of NH₄⁺ = (0.04290 + Y)/0.110 M

From Henderson-Hasselbalch equation,

pH = pK_a + log[base]/[acid]

or, 9.00 = 9.26 + log {(0.02915 - Y) x 0.110}/{(0.04290 + Y) x 0.110}

or, log (0.02915 - Y) /(0.04290 + Y) = - 0.26

or, (0.02915 - Y) /(0.04290 + Y) = 0.5495

or, (0.02915 - Y) = 0.02357 + 0.5495Y

or, 1.5495Y = 0.00558

or, Y = 0.00360

Thus the moles of HCl = 0.00360 mol

Molar mass of HCl = 36.5 g/mol

Therefore, the mass of HCl = 36.5 x 0.00360 = 0.131 g

Therefore, this buffer neutralizes 0.131 g of HCl before the pH falls below 9.00.