Question

Mix 40 drops 1 M NH,CI (ag) and 40 drops 1 M NaOH (ag) in your reqular test tube. Mix well and observe the reaction. Transfer the solution to a shallow well plate, Use Universal Indicator to determine the estimated pH of the reaction mixture observations: Estimated pH of Reaction Mixture: Products NaOH (aq) NH,CI (aq) H0 NaOHa) NHACla Balc'd Molc Eq: tci NHacp Na'a lonic Eq: NH + He Ore) +NHa (an + OH caa) Net lonic Eq: Calculation: Assume you combined 40.0 mL each of 1.0 M NH.CI and 1.0M NaOH. Calculate the theoretical pH of the resulting solution. How does your calculated value compare to the estimated pH of your reaction mixture? C E

Answer 1

The total volume of the mixture is ( 40 + 40) mL = 80 mL

Concentration of NH₄Cl in the mixture = 1.0 M * ( 40/ 80) = 0.5 (M)

Concentration of NaOH in the mixture = 1.0 M * ( 40/ 80) = 0.5 (M)

The reaction is:

                                                        NH₄⁺ (aq)    +    OH^- (aq)    =     H₂O (l)     +       NH₃ (aq)

            I :                                          0.5 M              0.5 M                  -                      0 M

            C:                                         - 0.5 M            - 0.5 M            -                  + 0.5 M

          E:                                           0                        0        - 0.5 M

[NH₃] = 0.50 (M)

NH₃ is a weak base with K_b = 1.8 * 10^-5 . NH₃ dissolved in water dissociates weakly to produce OH^-.

[OH^-] = ( K_b * [NH₃] ) ^1/2 = ( 1.8 * 10^-5 * 0.5) ^1/2 (M) = 0.003 (M)

pOH = - log[OH-] = - log ( 0.003) = 2.52

pH = 14 - pOH = 14 - 2.52 = 11.48