Question

An aqueous solution of phosphoric acid, that is 0.500 percent by weight phosphoric acid, has a density of 1.0010 g/mL. A student determines that the freezing point of this solution is -0.129 °C.

Based on the observed freezing point, what is the percent ionization of the acid and the value of K_a1? Assume that only the first ionization of the acid is important.
K_f for H₂O is 1.86 °C/m.

% ionized	=    %
Ka1	=

Answer 1

Answer:-

Given:-

freezing point of pure water (H₂O) T_f⁰ = 0 ⁰C

freezing point of solution T_{f = -} 0.129 ⁰C

decrease in freezing point  T_f  = T_f⁰ - T_f = 0 - (- 0.129) = 0.129 ⁰C

freezing point depression constant of water (H₂O) (Kf) = 1.86 ⁰C /m

density of phosphoric solution = 1.0010 g / ml

wt. of phosphoric acid (H₃PO₄) by (w/v) = 0.500

molar mass of phosphoric acid (H₃PO₄) = 98 g mol^-1

As we know that 0.500 percent by weight of phosphoric acid (H₃PO₄) means that 0.500 g of phosphoric acid (H₃PO₄) is present in 100 g of aqueous solution.

the wt. of phosphoric acid (H₃PO₄) = 0.500 g

wt of water (H₂O) = 100 - 0.500 = 99.5 g = 0.0995 Kg

therefore

no. of moles of phosphoric acid (H₃PO₄) (n) = wt. of phosphoric acid (H₃PO₄) / molar mass of phosphoric acid (H₃PO₄)

no. of moles of phosphoric acid (H₃PO₄) (n) = 0.500 g / 98 g mol^-1

no. of moles of phosphoric acid (H₃PO₄) (n) = 0.0051 mol

As we know that

molaity of phosphoric acid (H₃PO₄) solution (m) = no. of moles of phosphoric acid (H₃PO₄) (n) / wt of water (H₂O) in Kg

molaity of phosphoric acid (H₃PO₄) solution (m) = 0.0051 mol / 0.0995 Kg

molaity of phosphoric acid (H₃PO₄) solution (m) = 0.05126 m

According to the formula

T_f  = i Kf m

0.129 ⁰C = i 1.86 ⁰C /m 0.05126 m

Van't Hoff factor (i) =  0.129 ⁰C / 1.86 ⁰C /m 0.05126 m

Van't Hoff factor (i) =   0.129 / 0.0953436

Van't Hoff factor (i) = 1.353 ----------------------------------------------- (1)

As we know that

mass of solution = volume of solution density of solution

volume of solution phosphoric acid (H₃PO₄) = mass of (H₃PO₄) solution /  density of (H₃PO₄) solution

volume of solution phosphoric acid (H₃PO₄) = 100 g / 1.0010 g / ml

volume of solution phosphoric acid (H₃PO₄) = 99.9 ml = 0.0999 L

Also we know that

Molarity of phosphoric acid (H₃PO₄) = no. of moles of (H₃PO₄) (n) / volume of solution (H₃PO₄) in Liter

Molarity of phosphoric acid (H₃PO₄) =  0.0051 mol / 0.0999 L

Molarity of phosphoric acid (H₃PO₄) = 0.0512 M

As we know that phosphoric acid (H₃PO₄) ionized as follows:-

   H₃PO₄       H⁺ + H₂PO₄^-

Initial 0.0512 M 0 0   

Change - + +

Equilibrium (0.0512 M - )    + +

since is the degree of dissociation of phosphoric acid (H₃PO₄).

therefore

total moles of equilibrium = (0.0512 M - ) + + = 0.0512 M +

Also we know that

Van't Hoff factor (i) = 0.0512 M + /  0.0512 M ---------------------------------- (2)

from equation no. 1 and 2 we get

1.353 = 0.0512 M + /  0.0512 M

0.0512 M +   = 0.0512 M    1.353

0.0512 M +   = 0.0692736 M

= 0.0692736 M - 0.0512 M

= 0.01807

therefore

percent ionization of  phosphoric acid (H₃PO₄) (%) = degree of dissociation of phosphoric acid ()   100

percent ionization of  phosphoric acid (H₃PO₄) (%) = 0.01807 ​​​​​​​ 100

percent ionization of  phosphoric acid (H₃PO₄) (%) = 1.807 % ( i.e the answer)

As we know that

H₃PO₄    H⁺ + H₂PO₄^-

Initial 0.0512 M 0 0   

Change - 0.01807 M +0.01807 M +0.01807 M

Equilibrium (0.0512 M - 0.01807 M )    +0.01807 M +0.01807 M

0.03313 M +0.01807 M +0.01807 M
therefore we get

molar concentration of [H₃PO₄]= 0.03313 M

molar concentration of [H⁺ ] = 0.01807 M

molar concentration of [H₂PO₄^-] = 0.01807 M

As we know that

equilibrium constant of H₃PO₄ (K₁) = [H⁺ ][H₂PO₄^-] / [H₃PO₄]

equilibrium constant of H₃PO₄ (K₁) =  0.01807   0.01807 / 0.03313  

equilibrium constant of H₃PO₄ (K₁) =  0.0003265249 / 0.03313  

equilibrium constant of H₃PO₄ (K₁) = 0.009856

equilibrium constant of H₃PO₄ (K₁) = 9.856 10^-3 ( i.e the answer)