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In: Chemistry
Q-) A 25 ml of diluted bleach solution was titrated with 19.5 ml of 0.3 M...
Q-) A 25 ml of diluted bleach solution was titrated with 19.5 ml of 0.3 M of sodium thiosulfate (Na2S2O3), if the dilution factor = 5, then the % of NaOcl in org. bleach equals ( density of original. bleach= 1.08 g/ml) ??
Solutions
queen_honey_blossom answered 1 year agoRelated Solutions
A 25 mL aliquot of an HCl solution is titrated with 0.100 M NaOH. The equivalence...
A 25 mL aliquot of an HCl solution is titrated with 0.100 M
NaOH. The equivalence point is reached after 21.27 mL of the base
were added. Calculate 1) the concentration of the acid in the
original solution, 2) the pH of the original HCl solution and the
original NaOH solution, 3) the pH after 10.00 mL of NaOH have been
added, 4) the pH at the equivalence point, and 5) the pH after
25.00 mL of NaOH have been...
A 10.0 mL sample of liquid bleach is diluted to 100. mL in a volumetric flask....
A 10.0 mL sample of liquid bleach is diluted to 100. mL in a
volumetric flask. A 25.0 mL aliquot of this solution is analyzed
using the procedure in this experiment. If 10.4 mL of 0.30 M
Na2S2O3 is needed to reach the equivalence point, what is the
percent by mass NaClO in the bleach? Hint: Percent by mass is grams
of NaClO / mass of bleach. Presume a density for the bleach of 1.00
g/mL.
A 20.0 mL solution of 0.1004 M KI is titrated with a 0.0845 M solution of...
A
20.0 mL solution of 0.1004 M KI is titrated with a 0.0845 M
solution of AgNO3. What are the voltages after adding 1.0, 15.0,
and 25.0 mL of AgNO3? Ksp = 8.3x10^-17
Jackie diluted 7.5 mL weak acid with 7.5 mL distilled water. Then she titrated the diluted...
Jackie diluted 7.5 mL weak acid with 7.5 mL distilled water.
Then she titrated the diluted weak acid with 0.100 M NaOH. It takes
22.52 mL of NaOH to reach the equivalence point. What is the
original concentration of the weak acid? What is the diluted
concentration of the weak acid?
. A 35.00-mL solution of 0.2500 M HF is titrated with a standardized 0.1532 M solution...
. A 35.00-mL solution of 0.2500 M HF is titrated with a
standardized 0.1532 M solution of NaOH at 25 °C. The Ka value for
HF is 6.8 x 10-4 . (a) What is the pH of the HF solution before
titrant is added? (b) How many milliliters of titrant are required
to reach the equivalence point? (c) What is the pH at 0.50 mL
before the equivalence point? (d) What is the pH at the equivalence
point? (e) What...
50.0 mL of a 0.100 -M HoAc solution is titrated with a 0.100 -M NaOH solution....
50.0 mL of a 0.100 -M HoAc solution is titrated with a 0.100 -M
NaOH solution. Calculate the pH at each of the following
points.
Volume of NaOH added: 0, 5, 10, 25, 40 ,45 , 50 , 55 , 60 , 70 ,
80 , 90 , 100
A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution....
A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M
KOH solution. Calculate the pH after the following additions of the
KOH solution: (a) 0.0 mL, (b) 5.0 mL, (c) 10.0 mL, (d) 12.5 mL, (e)
15.0 mL. (25 points)
A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution....
A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M
KOH solution. Calculate the pH after the following additions of the
KOH solution: (a) 0.0 mL, (b) 5.0 mL, (c) 10.0 mL, (d) 12.5 mL, (e)
15.0 mL. (25 points)
A 10.0 mL solution of 0.300 M NH3 is titrated with a 0.100 M HCl solution....
A 10.0 mL solution of 0.300 M NH3 is titrated with a
0.100 M HCl solution. Calculate the pH after the addition
of 20.0 mL of the HCl solution.
(potentially useful info: Ka of NH4+ = 5.6 x 10−10)
A 35.00 mL solution of 0.2500 M HF is titrated with a standardized 0.1083 M solution...
A 35.00 mL solution of 0.2500 M HF is titrated with a
standardized 0.1083 M solution of NaOH at 25 degrees C.
(a) What is the pH of the HF solution before titrant is
added?
(b) How many milliliters of titrant are required to reach the
equivalence point?
(c) What is the pH at 0.50 mL before the equivalence point?
(d) What is the pH at the equivalence point?
(e) What is the pH at 0.50 mL after the equivalence...
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