Question

Need help with 3 and 4 on calculations. some info is on top of the page

Data Sheet for Lab #10 Name: Partner: Data/Observations: Volume of (NH), SO, used: Suomi initial pH of (NH4)2SO4 6.79 Equivalence vol. of NaOH: 20.00 pH at half-equivalence: 10,63 Balanced Chemical Equation: 1. Write the balanced equation for the reaction of ammonium sulfate and sodium hydroxide. (NH4)2SO4(s)? 2 N.0H(1) Nasa, () + 2NH3(-) + 2H2O 2. Write the balanced equation showing the ammonium ion acting as an Arrhenius acid. NH₃ + (aq) + OH (y) - NH₃(aq) + H₂O (6) Calculations: Show calculations for the following values: 1. The concentration of the ammonium ions from its mass and total volume. 2. 2.679 (NH4)₂SO4 I mad (Nity), soy : 6,020ml 132. N, GH)2SO4 The concentration of the ammonium ions from the equivalence volume. 0.020mol (NH4)₂ Soy x 0 out & lx , 0.001 ml (NH4)2SO4 1000 The acid-dissociation constant, K, from the pH at half-equivalence. 3. The acid-dissociation constant, K, from the initial pH and concentration of the acid. Show the ICE table and all calculations. Discussion: 1. How well do these K, values agree with the value reported in the textbook? 2. What did you learn as a result of performing this experiment?

Answer 1

I am trying to solve your problem using values given by you.

Salution (3) we knouw that pH af aidic bufer can be calculated [Salt] LAdd pukat og at laalf quivalanc poiut Aeid Esalt salf] CAcid aener a nentoued in given dada pka z 1o.63 £9.97 2 logka10.63 ka -11 As the uacion equation NH) Sacs)2NAOM Nasoglog) +2NH3a) t 2,ae) at the intial theu ts The asid teization equation NH tH20 NH+ Hg

The molo of (NHa S0g 2.67 pa malon mass 132 1989 mole -2 z 2-02X10 mol 20.0202 mole ICE Tahle NHg faitrlT) chauge C) Equililain(E 00202- O0202 O giuen that inilial pH 679 Now Leg-679 2 162x10 TNH3J [M30J kaz .0262-K (r62XI0T) (1-62X161) ka o.0202-200202 0.020 2 hal 50XI0 Lilsu

Kaz 162X62 X 4.09-X10 2 62 ka 1013 Ka 6.98X1014