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Anna [14]
2 years ago
8

What is the ph of a solution that has a [h ] = 0. 0039 m?

Chemistry
1 answer:
Sedbober [7]2 years ago
6 0

The pH of a solution that has a H⁺ ion concentration is 0.0039M is 2.4.

<h3>What is pH?</h3>

pH of any solution gives idea about the acidity or basicity of solution and it will be calculated as:

pH = -log[H⁺]

Given that, concentration of H⁺ ion = 0.0039 M

On putting tis value in pH equation, we get

pH = -log(0.0039)

pH = -(-2.4) = 2.4

Hence required value of pH is 2.4.

To know more about pH, visit the below link:
brainly.com/question/24595796

#SPJ1

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We can <u>find the equivalent number of O₂ molecules for 100 molecules of CO₂</u> using a <em>conversion factor containing the stoichiometric coefficients of the balanced reaction</em>, as follows:

  • 100 molecules CO₂ * \frac{6moleculesO_2}{4moleculesCO_2} = 150 molecules O₂

150 molecules of O₂ would produce 100 molecules of CO₂.

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Suppose that 0.00150 moles of CO2 (44.0 g/mol) effuse out of a pinhole in 1.00 hour. How many moles of N2 (28.0 g/mol) would eff
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Silver metal can be prepared by reducing its nitrate, AgNO3 with copper according to the following equation:
Lerok [7]

%yield = 88.5%

<h3>Further explanation</h3>

Given

Reaction

Cu(s) + 2 AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)

Required

The percent yield

Solution

mol AgNO₃(MW=169,87 g/mol) :

= mass : MW

= 127 : 169.87

= 0.748

mol Ag from equation :

= 2/2 x mol AgNO₃

= 2/2 x 0.748

= 0.748

Mass Ag (theoretical) :

= mol x Ar Ag

= 0.748 x 108

= 80.784

% yield = (actual/theoretical) x 100%

%yield = 71.5/80.784 x 100%

<em>%yield = 88.5%</em>

7 0
2 years ago
Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid
Phoenix [80]

Answer:

pH = 8.0

Explanation:

First, we have to calculate the moles of NaOH.

35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol

Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb)               <em>where Cb is the initial concentration of the base</em>

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

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