Question

please help with both of these problems!

Circle the correct K, that is used to find the Kai of Taratic Acid. Calculate the Ksi (4 pts) (Show work) Ka - 9.20 x 10 Ka:" 4.3 1 x 10-5 Calculate the activity coefficients for each ion of 0.0029 M Ca(POh. Hint: first calculate ionic strength. (Show work) (10 pts)

Answer 1

Q1.

For a generic diprotic acid like tartatic acid.

The reaction for first hydrolysis of the base is

$A^{2-} + H_2O \overset{K_{b1}}{\rightarrow} HA^- + OH^-$

The reaction for second dissociation of acid is

$HA^-+ H_2O \overset{K_{a2}}{\rightarrow} A^{2-} + H_3O^+$

Now adding the above two reactions we get

$\cancel{A^{2-}} + H_2O \overset{K_{b1}}{\rightarrow} \cancel{HA^-} + OH^-$

$\cancel{HA^-}+ H_2O \overset{K_{a2}}{\rightarrow} \cancel{A^{2-}} + H_3O^+$

$2H_2O \overset{K_w}{\rightarrow} H_3O^+ + OH^-$

Hence, $K_w = K_{b1} \times K_{a2}$

Hence, Ka2 is used to calculated Kb1.

Since,

Kw = 10-14

Kb1 can be calculated as

Kbl =-= Ku 10-14 Kg2 _ 4. 3 × 10-5-2.32 × 10-10

Q2.

Ionic strength can be calculated as

2 CİNİ

Where $C_i$ = concentration of i th ion.

$z_i$ = charge on i th ion.

In $Ca_3(PO_4)_2$, there are two types of ions, .

Ca2+ and Por

Concentration of solution = 0.0029 M

It dissociates as

$Ca_3(PO_4)_2 \rightarrow 3 Ca^{2+} + 2 PO_4^{3-}$

Hence, concentration of the cation = $C_+ = 0.0029 \ M \times 3 = 0.0087 \ M$

Concentration of anion = $C_- = 0.0029 \ M \times 2 = 0.0058 \ M$

Charge on cation =

Charge on anion =

hence,

$I = \frac{1}{2} \sum C_i z_i^2 \\ \Rightarrow I = \frac{1}{2} \left ( C_+z_+^2 + C_-z_-^2 \right ) \\ \Rightarrow I = \frac{1}{2} \left ( 0.0087\ M \times 2^2 + 0.0058\ M \times (-3)^2 \right ) \\ \Rightarrow I = 0.0435\ M$

Now the activity coefficient can be calculated as

$\log \gamma_i = -A z_i^2 \sqrt{I}$

Where $\gamma_i$ = activity coefficient of i th ion.

A = 0.509 for aqueous solution at 25 C.

Hence, for $Ca^{2+}$ ion

$\log \gamma_+ = -0.509 \times 2^2 \sqrt {0.0435} = -0.425 \\ \Rightarrow \gamma_+ = 10^{-0.425} \approx 0.376$

for $PO_4^{3-}$

$\log \gamma_- = -0.509 \times (-3)^2 \sqrt {0.0435} = -0.955 \\ \Rightarrow \gamma_- = 10^{-0.955} \approx 0.111$

Hence, the activity coefficients are 0.376 and 0.111 approximately for Ca2+ and PO43- respectively.