Question

2. Calculate the pH to the 100th's place when 10mL of 0.890M ammonium iodide NH4I (pka=9.26) is titrated with 1.436 M sodium hydroxide to the equivalence point.

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Answer 1

Given: pka of NHAI : 9.26 Concentration of NHI 0.8901 Volume of NHA I = lomb =0.012 Concentration of NaOH = 1.436 M Saleebian :- The equivalence point is a point of males o acids is equiblent to moles of base of sumber the mooi Nhy I is asall of weak base (NH4OH) and strong acid (HI). It is acidic solcetion NHL • Ngô pka = 9.26 -pka Ka =10 - 9.26 Ka = 10 [Ka = 5.50x100 l A balanced chemical NH I + NaOH (ag) reaction Na Ist NH 3 3 + H20 (1) Neutral weak a strong Acidic Salt hour alt base

Neember of indes og NHL I = Moleky x volume =(0.890 melly) (0.014) Muit = 0.0089 mil RHLE Z Based on a balanced chemical equation, are mole of the areacts coith one molo of NaOH then , 0.0089 mol of NH AI reacts with 0.0089 mol of NACH Malarity of NaOH = Volume (2) NaOH = moles (mal) mol Neember of moles of NaOH = Molarity * volume 0.0089 -6.436 mol /-) * V V= 0.0089 mol 1.436 gol/L V = 0.0062 L = 6.2 ml One mole of NHI paoduce one mole of I NH₂ 0.0089 ml of NHAI produce 0.0089 mole of NH3

Nha ay) +4%) = Memo) + Pers) Weak base Concentaation - males of NH₃ Total volume 13 Total volume call +0.0062 h = 0.01622 0.0089 med Concentration 0.0162 L C =0.5494 M Как, = (xo 14 Kb = 1x1014 TX 10-14 5.50xjo le Kb = 18 10 Concentaation NH3 (ag) NHAT (ag) OH (9) Initial 0.5494M change T -x + x +0 Equilibrium (0.5494 - x

Dissociation constant of bass as the ratio of the product of the products to the product of the reactants each paired to coofficients Rx = [N44.* ] [ot] lexo5 å defined of concentration of concentration power of = 7.8 X/ 05 [NHg 1.8x10x (6.5494-x) to 0.5494. 2.0 is very small , zeplace (6.5494-06 x² = (1.8x103) (0.5494) I : 9.8892 X10 I x = 3.1447X103 (NH4+-[OH 7 - x = 3:14 47x 103 poh is the negative lapecithms of hydroxido con pOH = -log [õh - log (3: 1447 x103) - log 3.1447 +3 log 10 = -0.4976 +3

| POH= 2.50 pH + pOH = 14 pH = 14 - POH pH = 14-2.50 [ pH = 11.50 Ones tenths tens & 11.50 Hundredths decimal point d the Boletion pH is 11:50

pH of a solution is 11.50