Answer to Question #185454 in Chemistry for dsaa

Question

Answer and show all work for part a,b,c, and d (dont leave out any parts)

A new weak acid, HG, has been synthesized. The a titration of 0.115g of HG requires 16.00 mL of 0.10 M KOH solution to reach the equivalence point.

(a) Write the dissociation equation for HG in water and its equilibrium expression (K_a). 

(b) Write the net ionic equation for the titration reaction. 

(c) Calculate the molar mass of HG. 

(d) After 8.00 mL of the base is added, the pH of the solution is 6.20. What is the value for the K_a of HG? 

Answer 1

Solution:

HG is a weak monoprotic acid.

(a):

The dissociation equation for HG in water:

HG(aq) + H₂O(l) = H₃O⁺(aq) + G^-(aq)

or

HG(aq) = H⁺(aq) + G^-(aq)

The equilibrium expression (K_a):

(b):

The balanced molecular equation:

HG(aq) + KOH(aq) = KG(aq) + H₂O(l)

The complete ionic equation:

H⁺(aq) + G^-(aq) + K⁺(aq) + OH^-(aq) = K⁺(aq) + G^-(aq) + H₂O(l)

The net ionic equation for the titration reaction:

H⁺(aq) + OH^-(aq) = H₂O(l)

(c):

HG(aq) + KOH(aq) = KG(aq) + H₂O(l)

According to the equation above: n(HG) = n(KOH)

n(HG) = m(HG) / M(HG)

n(KOH) = C(KOH) × V(KOH)

Hence,

m(HG) / M(HG) = C(KOH) × V(KOH)

M(HG) = m(HG) / (C(KOH) × V(KOH))

M(HG) = (0.115 g) / (0.10 mol L^-1 × 0.016 L) = 71.875 g/mol

The molar mass of HG is 71.875 g/mol

(d):

HG(aq) + KOH(aq) = KG(aq) + H₂O(l)

n_o(HG) = m(HG) / M(HG) = (0.115 g) / (71.875 g/mol) = 0.0016 mol

According to the equation: n(HG) = n(KOH) = n(KG)

n(KOH) = C(KOH) × V(KOH) = 0.10 mol L^-1 × 0.008 mol = 0.0008 mol

n(KG) = n(KOH) = 0.0008 mol

n(HG) = n_o(HG) - n(KOH) = 0.0016 mol - 0.0008 mol = 0.0008 mol

The Henderson-Hasselbach equation:

pH = pK_a + log([KG]/[HG])

pH = pK_a + log(n(KG)/n(HG))

6.20 = pK_a + log(0.0008 / 0.0008)

pK_a = 6.20

K_a = 10^-6.20 = 6.31×10^-7

K_a = 6.31×10^-7