You are making 2 L of a formic acid (HCOOH) sodium formate (HCOONa) buffer solut

Question

You are making 2 L of a formic acid (HCOOH) sodium formate (HCOONa) buffer solution with a pH of 3.7. At your disposition you have: A sodium formate powder A solution that is 99% by mass formic acid and has a density of 1.2 g/cm^3. If the concentration of formic acid in the buffer is 0.20 M, how many grams of sodium formate and how many milliliters of formic acid solution will you use? You have to identify an unknown monoprotic acid. To do that, you take 0.211g of the acid and then titrate it with a 0.0500M solution of NaOH. The pH of the solution after you added 33.64 mL of NaOH is 4.89. The equivalence point is reached after you added 44.70 mL of the NaOH solution. What is the molar mass of the unknown acid? What is the pKa of the acid? Consider the following equilibrium at 800 K, C(s) + H_2O(g) reversible CO(g) + H_2(g) delta H degree = +131 kJ Kc = 2.7 times 10^-2 Determine if the amount of CO(g) increases, decreases or remains unchanged under the following conditions: The temperature increases: The volume of the container decreases: Determine if the amount of H2(g) increases, decreases or remains unchanged under the following conditions: At 800 K, [H_2O]= 0.2M [CO] =0.3M [H_2] = 0.5M with solid carbon present in the container. At 800 K, [H_2O] = [CO] and [H2] = 0.01M with solid carbon present in the container.

Explanation / Answer

pH of a buffer is related with its concentration by Hinderson hasselbalch equation.

pH = pka + log [formate/formic acid]

or, 3.7 = 3.75 + log [formate/0.2]

or, [formate] =0.178 M

0.178 M = 0.178 moles/L. So, in 2 L number of moles of sodium formate = 2*0.178 =0.356 moles

Amount of sodium formate = 0.356 moles *68gm/mol =24.21 gm

[formic acid] = 0.2 M = 0.2 moles/L. 2 L solution contain 0.4 moles formic acid

formic acid is 99% by mass. this means 100gm formic acid solution contains 99 gm formic acid.

volume of 100gm solution = 100gm/1.2 gmcm^-3 =83.33 mL

99 gm formic acid = 99gm/46gmmol^-1 = 2.25 moles

2.25 moles formic acid is present in 83.33 ml solution. So, 0.4 moles are present in

83.33mL *0.2moles/2.25moles= 7.41 mL

Hence, 7.4 mL formic acid is required to make the buffer.

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