It is given that, [NH_{3}]_{0} [H^{+}]_{0} = [NH_{3}]_{x} [H^{+}]_{x} (equation 1) where subscript 0 refers to the initial values in a stock solution of NH_{3}, and the subscript x refers to the values in an unknown mixture (i.e during titration) both in 2M NH_{4}NO_{3}. Since pX is defined as the negative log_{10} of X, equation (1) can be written as:

p(NH_{3})_{0} + pH_{0} = p(NH_{3})_{x} + pH_{x} (equation 2)

p(NH_{3})_{x} = p(NH_{3})_{0} + pH_{0} – pH_{x} (equation 3) where pH_{x} is the measured pH at any time during titration. Thus, concentration of free NH_{3} can be calculated directly by applying equation 3. The total Cu^{2+} in solution at any point is defined as c_{cu} where

c_{cu} = [Cu^{2+}] + [Cu(NH_{3})^{2+}] [Cu(NH_{3})_{2}^{2+}] + [Cu(NH_{3})_{3}^{2+}] + [Cu(NH_{3})_{4}^{2+}]

The fractions, a_{n} of each species present at any point can be found by using the expressions for the stepwise formation constants.

a_{0} = [M]/c_{cu} = (1+ K_{1}[NH_{3}] + K_{1}K_{2}[NH_{3}]^{2} + K_{1}K_{2}K_{3}[NH_{3}]^{3})^{-1}

a_{1} = [Cu(NH_{3})^{2+}] / c_{cu} = K_{1}[NH_{3}]a_{0}

a_{2} = [Cu(NH_{3})_{2}^{2+}] / c_{cu} = K_{1}K_{2}[NH_{3}]^{2}a_{0}

The average number of NH_{3} bound per Cu^{2+} or the ligand number (n_{av}) can be calculated:

n_{av} = [NH_{3}]_{bound} / c_{cu} = ([NH_{3}]_{total} – [NH_{3}]_{free}) / c_{cu} = a_{1}+2a_{2+}3a_{3}+4a_{4}

In the experiment, the 0.7495 g (3 mmol) copper sulfate was dissolved in 100mL 2M NH_{4}NO_{3} solution to form 0.030 M Cu^{2+} solution (100 mL). 50 mL of copper solution was transferred to 200 mL beaker and titrated with 2M NH_{4}NO_{3} solution from a 50mL burette in 2 mL aliquots. The pH was measured after each addition. Given the information, calculate

Vol.of ammonia solution added | total volume of solution | Moles of ammonia added (mmol) | Measured pH | p(NH_{3}) |
[NH_{3}] calculated from p(NH_{3}) |
Moles of free ammonia (mmol) | Moles bound ammonia (mmol) | N_{av} |

2.00mL | 52mL | 4.55 | ||||||

4.00 mL | 54mL | 4.89 |

**Calculate** Moles of ammonia added (mmol), p(NH_{3}), [NH_{3}] calculated from p(NH_{3}), Moles of free ammonia (mmol), Moles bound ammonia (mmol), N_{av}

If the Moles of ammonia added (mmol) cannot be calculated, pretend that 1st value of moles of ammonia added was 0.384 mmol.

Thank you very much

## Expert Answer

To know what is the volume of ammonia added we know that the initial concentration is 2M [mmol/mL]

p(NH3)x = p(NH3)0 + pH0 – pHx to stimate P(NH3)x

(NH3)= 10^(-p(NH3)x)

ccu = [Cu2+] + [Cu(NH3)2+] [Cu(NH3)22+] + [Cu(NH3)32+] + [Cu(NH3)42+]

a0 = [M]/ccu = (1+ K1[NH3] + K1K2[NH3]2 + K1K2K3[NH3]3)-1

a1 = [Cu(NH3)2+] / ccu = K1[NH3]a0

a2 = [Cu(NH3)22+] / ccu = K1K2[NH3]2a0

nav = [NH3]bound / ccu = ([NH3]total – [NH3]free) / ccu = a1+2a2+3a3+4a4