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frosja888 [35]
2 years ago
11

A 1. 3×10−6mol sample of sr(oh)2 is dissolved in water to make up 25. 0 ml of solution. what is the ph of the solution at 25. 0∘

c?
Chemistry
1 answer:
VMariaS [17]2 years ago
4 0

The pH of the solution at 25 degree celsius of 1.3 × 10⁻⁶ moles of a sample of Sr(OH)₂ is 10.02.

<h3>How do we calculate pH?</h3>

The pH of any solution gives an idea about the acidic and basic nature of the solution and the equation of pH will be represented as:

pH + pOH = 14

Given that,

Moles of Sr(OH)₂ = 1.3 × 10⁻⁶ mol

Volume of solution = 25mL = 0.025L

The concentration of Sr(OH)₂ in terms of molarity = 1.3×10⁻⁶/0.025

                                                                                    = 5.2×10¯⁵M

Dissociation of Sr(OH)₂ takes place as:

                               Sr(OH)₂ → Sr²⁺ + 2OH⁻

From the stoichiometry of the reaction 1 mole of Sr(OH)₂ produces 2 moles of OH⁻.

Given that the base is a strong base and that it entirely dissociates into its ions, the hydroxide ion concentration is 5.2×10¯⁵×2 = 1.04×10¯⁴ M.

pOH = -log[OH⁻]

pOH = -log(1.04×10¯⁴)

pOH = 3.98

Now we put this value on the first equation we get,

pH = 14 - 3.98 = 10.02

Therefore, the value of pOH is 10.02.

Learn more about pH here:

brainly.com/question/24595796

#SPJ4

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2 years ago
The total volume required to reach the endpoint of a titration required more than the 50 mL total volume of the buret. An initia
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Answer:

The endpoint volume is 50.52 ±  0.14 mL

Explanation:

In a titration always is necessary to subtract the blank volume to the titrant volume to obtain the real volume of the titrant. Thus in this case, the total endpoint volume is the sum of the initial volume delivered and the second volume delivered, minus the blank volume:

V = (49.16±0.06 mL) + (1.69±0.04 mL) - (0.33±0.04 mL)

V = (49.16 + 1.69 - 0.33) ± (0.06+0.04+0.04) mL

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It is necessary to consider the sum of the errors too.

7 0
3 years ago
Analysis of an athletes urine found the presence of a compound with a molar mass of 312 g/mol. How many moles of this compound a
rewona [7]
<h3>Answer:</h3>

= 5.79 × 10^19 molecules

<h3>Explanation:</h3>

The molar mass of the compound is 312 g/mol

Mass of the compound is 30.0 mg equivalent to 0.030 g (1 g = 1000 mg)

We are required to calculate the number of molecules present

We will use the following steps;

<h3>Step 1: Calculate the number of moles of the compound </h3>

Moles=\frac{mass}{molar mass}

Therefore;

Moles of the compound will be;

=\frac{0.030}{312g/mol}

      = 9.615 × 10⁻5 mole

<h3>Step 2: Calculate the number of molecules present </h3>

Using the Avogadro's constant, 6.022 × 10^23

1 mole of a compound contains 6.022 × 10^23  molecules

Therefore;

9.615 × 10⁻5 moles of the compound will have ;

= 9.615 × 10⁻5 moles × 6.022 × 10^23  molecules

= 5.79 × 10^19 molecules

Therefore the compound contains 5.79 × 10^19 molecules

5 0
3 years ago
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