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Greeley [361]
3 years ago
5

A 1.00 l sample of water contains 0.0036 g of cl- ions. determine the concentration of chloride ions in ppm if the density of th

e solution is 1.00 g/ml
Chemistry
1 answer:
Illusion [34]3 years ago
6 0
Parts-per-million is a form of concentration that is confusing to some. It is often used for aqueous solutions at low concentrations because, as you state in your question, in dilute solutions the density of water has not been affected ie. is still 1.00 g/mL.

parts per million is defined as micrograms of solute per milliliter of solution. This doesn't inherently make sense since micrograms are on the millionths scale, but mL are on the thousandth scale, so it would seem to be millionth per thousandth--Not so! Since the 1.00 mL of water weighs 1.00 g, ug/mL converts, for dilute aqueous solutions, to ug/g, which now makes sense as parts per million. So, from the information given, the calculation is quite simple:

0.0036 g Cl * 10^6 ug/1 g = 3600 ug Cl

1.00 L * 1000 mL/1L = 1000 mL

Concentration of Cl- in solution in ppm:

3600 ug Cl- / 1000 mL = 3.6 ppm Cl-
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A 25.0-mL sample of 0.150 M hydrocyanic acid is titrated with a 0.150 M NaOH solution. The Ka of hydrocyanic acid is 4.9 × 10-10
lara [203]

Answer:

The pOH = 1.83

Explanation:

Step 1: Data given

volume of the sample = 25.0 mL

Molarity of hydrocyanic acid = 0.150 M

Molarity of NaOH = 0.150 M

Ka of hydrocyanic acid = 4.9 * 10^-10

Step 2: The balanced equation

HCN + NaOH → NaCN + H2O

Step 3: Calculate the number of moles hydrocyanic acid (HCN)

Moles HCN = molarity * volume

Moles HCN = 0.150 M * 0.0250 L

Moles HCN = 0.00375 moles

Step 3: Calculate moles NaOH

Moles NaOH = 0.150 M * 0.0305 L

Moles NaOH = 0.004575 moles

Step 4: Calculate the limiting reactant

0.00375 moles HCN will react with 0.004575 moles NaOH

HCN is the limiting reactant. It will completely be reacted. There will react 0.00375 moles NaOH. There will remain 0.004575 - 0.00375 = 0.000825 moles NaOH

Step 5: Calculate molarity of NaOH

Molarity NaOH = moles NaOH / volume

Molarity NaOH = 0.000825 moles / 0.0555 L

Molarity NaOH = 0.0149 M

Step 6: Calculate pOH

pOH = -log [OH-]

pOH = -log (0.0149)

pOH = 1.83

The pOH = 1.83

6 0
3 years ago
Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.
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100%*(1/2^5)= 100%/32=3..125%
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What are the spectator ions in the equation below? Select all that apply.
Licemer1 [7]

Answer:

H+ and NO3- ions

Explanation:

The Pb²+ ions react with the SO4²‐ ions to form a solid precipitate, i.e. they bond together and undergo a phase change;

On the contrary, the H+ and NO3- ions are aqueous ions before the reaction and the same after the reaction, i.e. they don't change;

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