Question

In: Chemistry

1.687 g of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt (ACES–K , MW = 220.29 g/mol) is dissolved in...

1.687 g of N-2-acetamido-2-aminoethane-sulfonic acid potassium salt (ACES–K , MW = 220.29 g/mol) is dissolved in 90.76 mL of water. 15.18 mL of HCl is added to the solution, resulting in a pH of 7.25. Calculate the concentration of the HCl solution. The pKa of ACES is 6.85.

Expert Solution

no of moles of ACEs = W/G.M.Wt = 1.687/220.29 = 0.00765 moles

ACES + HCl --------------> HACES

I 0.00765 x 0

C -x -x +x

E 0.00765-x 0 +x

PH = PKa + log[HACES]/[ACES]

7.25 = 6.85 + logx/0.00765-x

7.25-6.85 = logx/0.00765-x

0.4 = logx/0.00765-x

logx/0.00765-x = 0.4

x/0.00765-x = 10^0.4

x/0.00765-x = 2.5118

x = 2.5118*(0.00765-x)

x = 0.01921-2.5118x

x+2.5118x = 0.01921

3.5118x = 0.01921

x = 0.01921/3.5118 = 0.0055moles

no of moles oc HCl = x = 0.0055moles

molarity of HCl = no of moles/volume in L

= 0.0055/0.01518 = 0.362 M >>>>>> answer

queen_honey_blossom answered 1 year ago

3.920 g of sodium acetate ( CH3COONa MW= 82.03 g/mol), is dissolved in 235.0ml of H2O...

3.920 g of sodium acetate ( CH3COONa MW= 82.03 g/mol), is dissolved in 235.0ml of H2O at 50 degree celsius (kw= K.48E-14). Calculate the pH of this soluion ( Ka= 1.63 E-5) Determine the % ionization

A 0.1358-g portion of potassium iodate (MW 214.00), about 2 g of potassium iodide, and 2...

A 0.1358-g portion of potassium iodate (MW 214.00), about 2 g of potassium iodide, and 2 mL of 6 M hydrochloric acid were dissolved in 25 mL of distilled water. The triiodide formed during the ensuring reaction was titrated to the starch end point with 31.94 mL of a thiosulfate solution. A 25.00-mL triiodide sample solution was titrated to the end point with 21.33 mL of the standardized thiosulfate solution. Calculate the concentrations of the thiosulfate solution and the triiodide...

Calculate the freezing point of a solution of 40.0 g methyl salicylate, (MW: 152.15 g/mol), dissolved...

Calculate the freezing point of a solution of 40.0 g methyl salicylate, (MW: 152.15 g/mol), dissolved in 800. g of benzene, C6H6. Kf of benzene is 5.10 C/m and the freezing point is 5.50°C for benzene.

A 0.7257 g mixture of KCN (MW = 65.116 g/mol) and NaCN (MW = 49.005 g/mol)...

A 0.7257 g mixture of KCN (MW = 65.116 g/mol) and NaCN (MW = 49.005 g/mol) was dissolved in water. AgNO3 was added to the solution, precipitating all the CN– in solution as AgCN (MW = 133.886 g/mol). The dried precipitate weighed 1.650 g. Calculate the weight percent of KCN and NaCN in the original sample.

1.567 g of an unknown Mohr’s salt was dissolved in 80 mL of the sulfuric acid...

1.567 g of an unknown Mohr’s salt was dissolved in 80 mL of the sulfuric acid / phosphoric acid mixture solution. This iron solution was then titrated with a potassium permanganate solution (0.02M), what will be the volume at the equivalence point for the titration? Show all your calculations including formulas.

A 100.0 mL solution containing 0.830 g of maleic acid (MW = 116.072 g/mol) is titrated...

A 100.0 mL solution containing 0.830 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.298 M KOH. Calculate the pH of the solution after the addition of 48.0 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. pH = At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated H2M, HM–, and M2–, which represent the fully...

A 100.0 mL solution containing 0.8366 g of maleic acid (MW = 116.072 g/mol) is titrated...

A 100.0 mL solution containing 0.8366 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.3203 M KOH. Calculate the pH of the solution after the addition of 45.00 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated H2M, HM–, and M2–, which represent the fully protonated, intermediate,...

A 100.0 mL solution containing 0.7590 g of maleic acid (MW = 116.072 g/mol) is titrated...

A 100.0 mL solution containing 0.7590 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.2906 M KOH. Calculate the pH of the solution after the addition of 45.00 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. pH=? At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated H2M, HM-, and M^2-, which represent the fully protonated,...

A 100.0 mL solution containing 0.753 g of maleic acid (MW = 116.072 g/mol) is titrated...

A 100.0 mL solution containing 0.753 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.259 M KOH. Calculate the pH of the solution after the addition of 50.0 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated H2M, HM–, and M2–, which represent the fully protonated, intermediate,...

A 100.0 mL solution containing 0.756 g of maleic acid (MW = 116.072 g/mol) is titrated...

A 100.0 mL solution containing 0.756 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.217 M KOH. Calculate the pH of the solution after the addition of 60.0 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. At this pH (pH = 9.48), calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated H2M, HM–, and M2–, which represent the...