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Phoenix [80]
3 years ago
7

Balance the following equations and then write the net ionic equations. (Use the lowest possible whole number coefficients. Incl

ude states-of-matter in your answers. Use instead of to balance equations. Write weak electrolytes in their undissociated form. Omit any coefficients of 1).
A. (NH4)3PO4(aq) + Bi(NO3)3(aq) + BiPO4(s) + NH4NO3(aq).
B. AGOH(s) + H2SO4 (aq) + Ag, SO4(s) + H2O.
C. HCIO2 (aq) + Mn(OH)2(s) + Mn(C102)2 (aq) + H2O.
Chemistry
1 answer:
Fiesta28 [93]3 years ago
8 0

Answer:

Explanation:

A )  (NH₄)₃PO₄(aq) + Bi(NO₃)₃(aq) =  BiPO₄(s) + 3 NH₄NO₃(aq).

B₁³⁺ +   PO₄⁻³  =  Bi ( PO₄ )

B )  AgOH(s) + H₂SO₄ (aq)  =  AgSO₄(s) + H₂O.

Ag⁺  + OH⁻ + H⁺ +  2 SO₄⁻² (aq)  =  Ag(SO₄)₂(s) + H₂O.

C )  HCIO₂ (aq) + Mn(OH)₂(s) =   Mn(Cl0₂)₂ (aq) + H₂O.

     H⁺  + OH⁻  =  H₂O .

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Quinine (C20H24N2O2) is the most important alkaloid derived from cinchona bark. It is used as an antimalarial drug. For quinine
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Answer:

The pH of saturated solution of the quinine is 10.05

Explanation:

Quinine (Q) is C20H24N2O2 has a molar mass of 324.4 g/mol

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pKb1 = 5.1

Step 1 : Calculate the Kb of Quinine

            pKb1 = - log [kb]

                5.1  = - log [kb]

                take Antilog of both side

             [kb] = 7.94 x 10∧-6

Step 2: Calculate the concentration of saturated solution of Q in mol/dm3

           From the question, 1900 ml of solution contains 1 g of Q

           Therefore,  1000 ml of solution will contain........... x g of Q

           x = 1000 /1900

           x = 0.526 g in 1 dm3

In calculating concentration in mol/dm3,

Concentration in mol/dm3 = concentration in g /dm3 divided by molar mass

Molar mass of Q = 324.4

Concentration in mol/dm = 0.526 /324.4

                                         = 0.0016 mol/dm3

Step 3: Calculating the Concentration of OH-

            At Equilibrium, Kb = x² / 0.0016

            7.94 x 10∧-6 = x² / 0.0016

            x = √ 0.0016 × 7.94 x 10∧-6

            x = 1. 127 × 10∧-4 mol/dm3

The concentration of OH- = 1. 127 × 10∧-4 mol/dm

Step 4:  Calculating the pH of Quinine

           Recall, pOH = - log [OH-]

           pOH = - log [1. 127 × 10∧-4]

           pOH = 3.948

           Also recall that pH + pOH = 14

           pH = 14 - 3.948

           pH = 10.05

           

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