Question

Volume NaOH added Moles of NaOH added pH (mL) 0 2.89 0 3.18 0.00010 0,00020 3.18 0.00030 3 3.42 3.56 4 0.00040 0.00050 5 3.67 Concentration of NaOH-0.1 M Concentration of HCHO-0.1 M 6 3.73 0,00060 0,00070 7 3.89 0,00080 8 3.99 9 4.07 0.00090 What is the pH at equivalence point? What is the volume of NaOH at equivalence point? a) 10 4.14 0.00100 b) Moles of NaOH at the equivalence point? Moles of HC:H0 at equivalence point? 11 4.21 0,00110 e) 12 4.27 4.33 4.42 0,00120 d) 13 0.00130 14 0,00140 e) Find Ka of HCaHsO2 Find pK. of HC:H0 15 4.47 0.00150 16 17 4.53 0,00160 0,00170 0.00180 4.59 18 4.66 19 4.72 4.80 0.00190 20 0,00200 21 4.87 0.00210 22 4.93 0.00220 23 24 0.00230 0.00240 4.99 5.08 0.00250 25 5.20 26 5.33 0,00260 0.00270 27 5.56 28 5.89 0.00280 29 6,69 998 0,00290 30 31 32 0.00300 10.1 0,00310 0.00320 10.18 33 10.21 0,00330 34 35 36 10.28 10.31 10.32 10.39 0,00340 000350 0,00360 37 0.00370 38 39 10.42 0.00380 0,00390 0.00400 10.48 10.5 40

Answer 1

The graph is made:

Volume NaOH pH 12 0 2.89 2 3.18 4 3.56 10 6 3.73 3.99 8 10 4.14 4.27 12 4.42 14 6 4.53 16 4.66 18 4 20 4.8 4.93 22 5.08 24 26 5.33 5.89 28 0 0 9.98 30 10 40 32 10.18 34 10.28 30 20 st

a) At the point of equilibrium, there is a pH of.

b) At the equivalence point, the volume of NaOH is 29 mL.

c) At the equivalence point, 0.0029 moles of NaOH were added.

d) The moles of weak acid at the equivalence point are equal to those of NaOH, 0.0029 mol.

e) The pH is read at the midpoint of equivalence, at 14.5 mL, it is equal to the acid pKa:

pH = pKa = 4.45

The Ka is calculated:

Ka = 10 ^ -pKa = 10 ^ -4.45 = 3.55x10 ^ -5

f) The pKa = 4.45

If you liked the answer, please rate it in a positive way, you would help me a lot, thank you.