We have 100.0 mL of 1.0 M formic acid, HCOOH solution. We determine the millimol

Question

We have 100.0 mL of 1.0 M formic acid, HCOOH solution. We determine the millimoles of HCOOH in the solution as (100.0 mL)*(1.0 M) = 100.0 mmole.

We are preparing a 1:1 buffer by adding solid NaOH as per the reaction below.

HCOOH (aq) + NaOH (s) ------> HCOONa (aq) + H2O (l)

We have a 1:1 buffer; hence, the millimoles of HCOOH and HCOONa in the buffer must be equal. This is possible only when HCOOH is half neutralized, i.e, we will have 50.0 mmole of HCOOH and 50.0 mmole of HCOONa in the prepared buffer.

We add 30.0 mL of 1.0 M HCl to the prepared buffer; the millimoles of HCl added = (30.0 mL)*(1.0 M) = 30.0 mmole.

HCl reacts with HCOONa in the buffer to form HCOOH; therefore, we have,

HCOONa (aq) + HCl (aq) -------> HCOOH (aq) + NaCl (aq)

Moles of HCl added = moles of HCOONa neutralized = moles of HCOOH formed.

Since we add 30.0 mmole HCl, mmoles of HCOONa neutralized = mmoles of HCOOH formed = 30.0 mmole.

Millimoles of HCOOH at equilibrium = (50.0 + 30.0) mmole = 80.0 mmole; millimoles of HCOONa at equilibrium = (50.0 – 30.0) mmole = 20.0 mmole.

Since the volume of the solution stays constant at (100.0 + 30.0) mL = 130.0 mL, we can express the ratio of the molar concentrations of HCOONa and HCOOH in terms of the number of millimoles of each, i.e,

[HCOONa]/[HCOOH] = (20.0 mmole)/(80.0 mmole) = 0.25.

Use the Henderson-Hasslebach equation.

pH = pKa + log [HCOONa]/[HCOOH]

====> pH = 3.75 + log (0.25) = 3.75 + (-0.6020) = 3.148 3.14.

(a) is the correct answer.

Explanation / Answer

10. In expcriment 4, some of you were given a roughly 1.0 M solution of formic acid to titrate Suppose you decided to make a 1:1 buffer by starting with 100.0 mL of this solution and adding NaOH (s). Assume that the volume remains constant when this solid is added. If you then add 30.0 mL of 1.0 M HCt to your buffer, what is the pH of your solution now? C) 3.14 correct answer b. 3.34 c. 3.48 d. 3.59

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