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uysha [10]
3 years ago
14

If the [H₃O⁺] of a solution is 1.7 x 10⁻³ M, what is the pH of the solution?

Chemistry
1 answer:
vesna_86 [32]3 years ago
6 0

Answer:

The pH of the solution is 2.77

Explanation:

Data obtained from the question include:

Concentration of hydronium ion, [H₃O⁺] = 1.7 x 10⁻³ M

pH =?

The pH is define as the measure of concentration of Hydrogen ion [H⁺] or hydronium [H₃O⁺] in an acid or a base. Mathematically, it is expressed as:

pH = –Log [H₃O⁺]

With the above equation, the pH can be obtained as follow:

pH = –Log [H₃O⁺]

[H₃O⁺] = 1.7 x 10⁻³ M

pH = –Log 1.7 x 10⁻³ M

pH = 2.77

Therefore, the pH of the solution is 2.77

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In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe
const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = 10^{-9} m.

Hence, 0.283 nm = 0.283 \times 10^{-9} m

  • Formula for coulombic energy is as follows.

             U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}

where,   e = 1.6 \times 10^{-19} C

            \epsilon_{o} = 8.85 \times 10^{-12}

          U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}

                         = 1.423 \times 10^{-18} J

  • As 1 eV = 1.6 \times 10^{-19} J

So,       1 J = \frac{1 eV}{1.6 \times 10^{-19}}

Hence,    U = \frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}

                   = 8.9 eV

  • Also,   1 J = \frac{10^{-3} kJ}{6.022 \times 10^{23}mol}

                = 1.67 \times 10^{-27} kJ/mol

Therefore, U = 1.423 \times 10^{-18} J \times 1.67 \times 10^{-27} kJ/mol

                     = 2.37 \times 10^{-45} kJ/mol

7 0
3 years ago
Suppose you need to prepare 141.9 mL of a 0.223 M aqueous solution of NaCl. What mass of NaCl do you need to use to make the sol
Afina-wow [57]

Answer:

1.811 g

Explanation:

The computation of the mass need to use to make the solution is shown below:

We know that molarity is

Molarity = \frac{Number\ of\ moles}{Volume\ in\ L}

So,

Number\ of\ moles = Molarity\ \times Volume\ in\ L

= 0.223\times 0.141

= 0.031 moles

Now

Mass = moles \times Molecualr\ weight

where,

The Molecular weight of NaCl is 58.44 g/mole

And, the moles are  0.031 moles

So, the mass of NaCL is

= 0.031 \times 58.44

= 1.811 g

We simply applied the above formulas

3 0
2 years ago
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