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DENIUS [597]
3 years ago
14

What volume of a concentrated HCl solution, which is 36.0% HCl by mass and has a density of 1.179 g/mL, should be used to make 5

.20 L of an HCl solution with a pH of 1.70
Chemistry
1 answer:
charle [14.2K]3 years ago
8 0

Answer:

The volume we need is 8.89 mL

Explanation:

We analyse data:

36 % by mass → 36 g of HCl in 100 g of solution

Solution's density = 1.179 g/mL

5.20L → Volume of diluted density

pH = 1.70 → [H⁺] = 10⁻¹'⁷⁰ = 0.0199 M

 HCl    →       H⁺     +    Cl⁻

                 0.0199

pH can gives the information of protons concentrations, so as ratio is 1:1, 0.0199 M is also the molar concentration of HCl

Let's verify the molar concentration of the concentrated solution:

We convert the mass to moles: 36 g / 36.45 g/mol = 0.987 moles

As the solution mass is 100 g, we apply density to find out the volume:

Density = Mass / volume → Volume = Mass / Density

Volume = 100 g / 1.179 g/mL → 84.8 mL

Let's convert the volume from mL to L in order to define molarity

84.8 mL . 1L/ 1000mL = 0.0848 L

Molarity → 0.987 mol / 0.0848L = 11.6M

Let's apply the dilution formula:

M concentrated . V concentrated = M diluted . V diluted

11.6 M . V concentrated = 0.0199M . 5.20L

V concentrated = (0.0199M . 5.20L) / 11.6M → 8.89×10⁻³L

We can say, that the volume we need is 8.89 mL

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formic acid buffer containing 0.50 M HCOOH and 0.50 M HCOONa has a pH of 3.77. What will the pH be after 0.010 mol of NaOH has b
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Answer:

pH = 3.95

Explanation:

It is possible to calculate the pH of a buffer using H-H equation.

pH = pka + log₁₀ [HCOONa] / [HCOOH]

If concentration of [HCOONa] = [HCOOH] = 0.50M and pH = 3.77:

3.77 = pka + log₁₀ [0.50] / [0.50]

<em>3.77 = pka</em>

<em />

Knowing pKa, the NaOH reacts with HCOOH, thus:

HCOOH + NaOH → HCOONa + H₂O

That means the NaOH you add reacts with HCOOH producing more HCOONa.

Initial moles of 100.0mL = 0.1000L:

[HCOOH] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOOH

[HCOONa] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOONa

After the reaction, moles of each species is:

0.0500moles HCOOH - 0.010 moles NaOH (Moles added of NaOH) = 0.0400 moles HCOOH

0.0500moles HCOONa + 0.010 moles NaOH (Moles added of NaOH) = 0.0600 moles HCOONa

With these moles of the buffer, you can calculate pH:

pH = 3.77 + log₁₀ [0.0600] / [0.0400]

<h3>pH = 3.95</h3>

3 0
3 years ago
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