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Anuta_ua [19.1K]
3 years ago
9

How much 10.0 M HNO3 must be added to 1.00 L of a buffer that is 0.0100 M acetic acid and 0.100 M sodium acetate to reduce the p

H to 5.05 ?
Chemistry
1 answer:
sladkih [1.3K]3 years ago
8 0
Initial moles of CH₃COOH = 1.00 x 0.0100 = 0.0100 mol

Initial moles of CH₃COONa = 1.00 x 0.100 = 0.100 mol


Let a be the volume of HNO₃ that must be added

Moles of HNO₃ added = a/1000 x 10.0 = 0.01a mol

CH₃COONa + HNO₃ => CH₃COOH + NaNO₃


Moles of CH₃COOH = 0.0100 + 0.01a

Moles of CH₃COONa = 0.100 - 0.01a


pH = pKa + log([CH₃COONa]/[CH₃COOH])

= pKa + log(moles of CH₃COONa/moles of CH₃COOH)


5.05 = 4.75 + log((0.100 - 0.01a)/(0.0100 + 0.01a))

log((0.100 - 0.01a)/(0.0100 + 0.01a)) = 0.3

(0.100 - 0.01a)/(0.0100 + 0.01a) = 10^0.3 = 2.0

0.03 a = 0.08

a = 2.67


<span>Volume of HNO3 = a = </span>2.13 mL
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3 years ago
A base has Kb of 2.5 x 10-11. which of the following statements is true?(1 point)
Furkat [3]
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3 0
3 years ago
• We obtained the above 10.00-mL solution by diluting a stock solution using a 1.00-mL aliquot and placing it into a 25.00-mL vo
garik1379 [7]

Answer:

a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) 0.0035 mole

c)  0.166 M

Explanation:

Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.

The equation of the reaction is expressed as:

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

1 mole         3 mole

The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b)  if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

10 ml            17.50 ml

(x) M              0.200 M

Molarity = \frac{0.2*17.5}{1000}

= 0.0035 mole

c) What was the molar concentration of phosphoric acid in the original stock solution?

By stoichiometry, converting moles of NaOH to H₃PO₄; we have

= 0.0035 \ mole \ of NaOH* \frac{1 mole of H_3PO_4}{3 \ mole \ of \ NaOH}

= 0.00166 mole of H₃PO₄

Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:

Molar Concentration =  \frac{mole \ \ of \ soulte }{ Volume \ of \ solution }

Molar Concentration = \frac{0.00166 \ mole \ of \  H_3PO_4 }{10}*1000

Molar Concentration = 0.166 M

∴  the molar concentration of phosphoric acid in the original stock solution = 0.166 M

6 0
3 years ago
Calculate the molarity of a solution prepared by dissolving 1.495 moles of lioh in enough water to give a final volume of 750. m
zhenek [66]
Molarity can be defined as the number of moles of substance dissolved in 1 L of solution.
In the given question ,
number of LiOH moles - 1.495 mol
Dissolved volume - 750 mL
molarity is calculated for 1 L = 1000 mL
In 750 mL - 1.495 mol of LiOH is dissolved
Therefore in 1000 mL - 1.495 mol / 750 mL x 1000 = 1.99 mol

4 0
3 years ago
At PO2 = 40 mm Hg, which statement about the saturation of either myoglobin (Mb) or hemoglobin (Hb) is true? At PO2 = 40 mm Hg,
Damm [24]

Answer:

At this partial pressure of oxygen, Mb would be almost completely saturated but Hb would not.  

Explanation:

The oxygen saturation curves for Mb and Hb are quite different. The curve for Mb is hyperbolic while that for Hb is sigmoidal.

Mb reaches oxygen saturation before Hb.

Thus, at a partial pressure of 40 mmHg, Mb is almost completely saturated but Hb is not.

4 0
3 years ago
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