Question

We’re going to titrate formic acid with the strong base, NaOH. There is initially 100. mL of 0.50 M formic acid and the concentration of NaOH is 1.0 M.

What is the initial pH of the formic acid solution?

2) What is the percent ionization under initial conditions?

3) After the addition of 10 mL of NaOH, what is the pH?

4) After the addition of 25 mL of NaOH, what is the pH? Think about where in the titration this brings you.

5) What volume of NaOH is required to reach the equivalence point?

6) What is the pH at the equivalence point?

7) What is the pOH at the equivalence point?

8) If, instead of NaOH being added, 0.05 moles of HCl is added by bubbling the gas through the solution. Assume that the volume has not changed. What is the percent dissociation of formic acid?

Could somone please help me and show your work please. Thank you so much

Answer 1

You need to analize the reaction of formic acid at equilibrium:

$HFor\rightarrow H^++For^-$

An construct a ICE table:

	$HFor\rightarrow$	$H^++$	$For^-$
Initial	0.5 M
Change	-X	+X	+X
Equilibrium	1-X	X	X

Using the mass action equation:

$Ka=\frac{[For^-][H^+]}{[HFor]}$

Ka for formic acid is 10^-3.77

Using the concentrations at equilibrium :

$Ka=\frac{X*X}{0.5-X}$ <----this yields a second degree equation:

$X^2+KaX-Ka=0$

$X^2+(10^{-3.77})X-10^{-3.77}(0.5)=0$ <---- you can solve this using the quadratic formula:

$X=9.13\times10^{-3}$ <----concentration of H+

$[H^+]=0.0129M$

$pH=-log[H^+]=-log(9.13\times10^{-3})=2.04$

and also the concentration for formate is 9.13x10^-3M

The percentage of dissociation is:

$100\times\frac{[For^-]}{[HFor]_o}\frac{formate\;at\;equilibrium}{initial\;concentration}$

$100\times\frac{9.13\times10^{-3}M}{0.5M}=1.83\%$

3)

After adding 10mL of NaOH:

You will be adding 0.01L*1mol/L=0.01mol of OH

$HFor+OH^-\rightarrow H_2O+For^-$

Since initially you the acid is dissociated less than 2 % you can assume the whole initial concentration is HFor, thus the initial moles of HFor are:

0.100L*0.5mol/L=0.05mol HFor

after adding the OH the acid remaing is:

$0.05mol-0.01mol=0.04mol$

and the moles of base formed are:

Using the henderson hasselbach equation:

$pH=pKa+log\left ( \frac{molFor^-}{molHFor} \right )$

$pKa=-log(Ka)=-log(10^{-3.77})=3.77$ <----this is from the definition of pKa

$pH=3.77+log\left ( \frac{0.01}{0.04} \right )=3.17$

4)

after addign 25mL of NaOH

at this point the moles of OH added are:

0.025*1M=0.025mol<-----this is half of the initilal moles of acid (0.5M*0.1L=0.05mol)

Hence you will have halaf the concentration of For^- and half the concentration of HFor

at this point the pH is the pKa=3.77

you can prove it using the henderson hasselbach equation

$pH=3.77+log\left ( \frac{0.05-0.25}{0.025} \right )=3.77$

5)

The volume required to reach the equilvalence point is:

Total moles of initial HFor=0.5*0.1=0.05mol

From the stoichiometry of the reaction above you need 0.05mol of NaOH to titrate the acid, thus:

$\frac{0.05mol}{1mol/L}=0.05L---->50mLNaOH$

6)

From the reaction

$HFor\rightarrow H^++For^-$

The equivalence point is where all the Hfor has been consumed and the pH is given by the base formed:

you need to analize the dissociation at equilibrium:

$For^-+H_2O\rightarrow HFor+OH^-$

Using an ICE table:

	$For^-$	$H_2O\rightarrow$		$OH^-$
Initial	1 M
Change	-X		+X	+X
Equilibrium	1-X		X	X

in this case you need to use the Kb:

$Kb=\frac{K_w}{K_a}=\frac{10^{-14}}{10^{-3.77}}=10^{-10.23}$

and using the mass action equation:

$Kb=\frac{[HFor][OH^-]}{[For^-]}$

and the concentrations at equilibrium:

$Kb=\frac{X*X}{0.5-X}$ ----->as above you need to rearrange thi equation:

$X^2+KbX-0.5Kb=0$

$X^2+(10^{-10.23})X-0.5(10^{-10.23})=0$

$X=5.43\times10^{-6}$ <--- this is the concentration of OH:

$[OH^-]=5.43\times10^{-6}M$

$pOH=-log[OH^-]=-log(5.43\times10^{-6})=5.26$

$pH=14-pOH=14-5.26=8.74$

7) pOH=5.26

8)

if you add 0.05mol of HCl, the concentration in the 100ml of H+ will be:

$\frac{0.05mol}{0.1L}=0.5mol/L$

you can make another ICE table, but in this case the initial H concentration is 0.5

	$HFor\rightarrow$	$H^++$	$For^-$
Initial	0.5	0.5
Change	-X	+X	+X
Equilibrium	0.5-X	0.5+X	X

And the mass action equation with the concentrations at equilibrium:

$Ka=\frac{X(X+0.5)}{0.5-X}$

$X^2+X(0.5+Ka)-0.5Ka=0$

$X^2+X(0.5+10^{-3.77})-0.5(10^{-3.77})=0$

$X=1.697\times10^{-4}$ <----ffrom the ICE table this is the concentration of For-

$[H^+]=0.5+1.697\times10^{-4}\approx 0.5M$

$pH=-log(0.5)=0.3$

The percentage of dissociation is:

$100\times\frac{[For^-]}{[HFor]_o}\frac{formate\;at\;equilibrium}{initial\;concentration}$

$100\times\frac{1.697\times10^{-4}}{0.5}=0.03\%$