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Concept explainers
Interpretation:
The pH buffer made from 475 mL of 0.200 M benzoic acid and 25 mL of 2.00 M NaOH with make 500 mL of 0.200 M HCOOH and 2.00 M NaOH volume has to be calculated.
Concept Introduction:
pH definition:
The concentration of hydrogen ion is measured using pH scale. The acidity of aqueous solution is expressed by pH scale.
The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.
pH = -log[H3O+]
Buffer solution:
- Buffer solution is defined as a solution that oppose changes in pH while adding little amount of either an acid or a base. In general, addition of acid or base does not affect the pH in buffer solution but if it is more than amount of conjugate base or conjugate acid, then buffer loses its buffering capacity.
- Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.
Henderson-Hasselbalch equation:
The pH at an intermediate stage of the titration is calculated by using the following equation,
pH = pKa + log [base][acid]where,pH = - log [H+]pKa = - log Ka (Ka - acid dissociation constant)
This equation is known as Henderson-Hasselbalch equation.
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Answer to Problem 19.106P
The required volume of the base NaOH is 36 mL and the volume of the formic acid HCOOH is 0.464= 464 mL.
Explanation of Solution
Consider the following
C6H5COOH(aq) + NaOH(aq) ⇄ Na2+ (aq) + C6H5COO−(aq) + H2O(l)
Record the given data,
Concentration of H−COOH = 0.200 M
Concentration of NaOH = 2.00 M
Volume of solution V= 500.0 mL
Concentration of benzoic acid = 0.200 M
Calculate the number of mole of benzoic acid present in the solutions,
Moles of C6H5COOH= (0.200 mol C6H5COOHL)(10−3 L1 mL)(475 mL)= 0.0950 mol of C6H5COOH
Calculate the number of mole of NaOH present in the solution,
Moles of NaOH = (2.00 mol NaOHL)(10−3 L1 mL) (25 mL)= 0.050 mol NaOH
NaOH is the limiting reagent,
Write the balance equation of the reaction,
C6H5COOH(aq) + NaOH(aq) ⇄ Na2+ (aq) + C6H5COO−(aq) + H2O(l)Initial (M): 0.0950 mol 0.050 mol − 0 −Change (M): −0.050 mol −0.050 mol − +0.050 mol −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−Equilibrium (M): 0.045 mol 0 mol − 0.050 mol −
The concentrations after the reaction are,
[C6H5COOH] = (0.045 mol C6H5COOH(475+ 25) mL)(1 mL10−3 L)= 0.090 M C6H5COOH
[C6H5COO−] = (0.050 mol C6H5COO−(475+ 25) mL)(1 mL10−3 L)= 0.10 M C6H5COO−
Calculate the pH of the reaction
Henderson-Hasselbalch equation is,
pH = pKa + log [base][acid]
pH = pKa + log (C6H5COO−C6H5COOH)
The Ka for benzoic acid is 6.3×10−5 (this value referred from Appendix table).
The pKa is −log(6.3×10−5) = 4.201
The pKa value is plugging above equation, to solve pH of the reaction,
pH = 4.201 + log (0.100.090)= 4.201 + log(1.1111)= 4.201 + 0.045757=4.24675pH=4.2
Calculate the formic acid (H-COOH)and use the Henderson-Hasselbalch equation,
pH = pKa + log [base][acid]
The Ka for formic acid is 1.8×10−4 (this value referred from Appendix table).
The pKa is −log(1.8×10−4) = 3.7447
pH = pKa + log (HCOO−HCOOH)
Benzoic acid pH value is 4.24675
Formic acid pKa value is 3.7447
The pH and pKa values are plugging above equation,
⇒4.24676 = 3.7447 + log (HCOO−HCOOH)⇒4.24676−3.7447= log (HCOO−HCOOH)
Hence,
log (HCOO−HCOOH) = 0.50206pKa value for formic acid is 3.177313 (HCOO−HCOOH) = 3.177313(HCOO−) = 3.177313 (HCOOH)
Since the conjugate acid and the conjugate base are in the same volume, the mole ratio and the molarity ratios are identical.
Moles of HCOO− = 3.177313 mol HCOOH
The total volume of the solution is,
=(500 mL)×(10−3 L1 mL)= 0.500 L
Let,
Va= Volume of acid solution added and
Vb= Volume of base solution added
Thus,
Va + Vb = 0.500 L
The reaction between the formic acid and the sodium hydroxide is,
HOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l)
The moles of NaOH added equal the moles of HCOOH reacted and the moles of HCOONa formed.
Moles NaOH = (2.00 mol NaOH/L) (Vb) =2.00 Vb molTotal moles HCOOH = (0.200 mol HCOOH/L)(Va)=2.00 Va mol
The stoichiometric ratios in this reaction are 1:1 ratio,
Moles HCOOH remaining after theraction = (0.200 Va−2.00 Vb) molMoles HCOO− = moles HCOONa = moles of NaOH =2.00 Vb
Using these moles and the mole ratio determined for the buffer solution gives,
Moles HCOO− = 3.177313 mol HCOOH2.00 Vb mol = 3.177313 (0.200 Va−2.00 Vb ) mol2.00 Vb = 0.6354626 Va− 6.354626 Vbadded for Vb_8.354626 Vb= 0.6354626 Va
The volume relationship given above calculation,
Va + Vb = 0.500 LVa= (0.500 −Vb)L
Calculate for the volume (V)
8.354626 Vb= 0.6354626 Va (0.500 −Vb)8.354626 Vb = 0.3177313−0.6354626 Vb
Added for volume (Vb)
8.9900886Vb = 0.3177313Vb= 0.31773138.9900886Vb= 0.0353424Vb= 0.035 L NaOH
Next calculate for volume (Va)
Va = 0.500− 0.0353424= 0.4646576=0.465 L HCOOH
Hence, the volume of NaOH is 0.500−0.464 = 0.036 L = 36 mL and the volume of the formic acid HCOOH is 0.464= 464 mL.
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