Calculate The Amount Of Heat Needed To Boil 132.g Of Water (H20), Beginning From A Temperature Of 7.4

Ai. The concentration of hydronium ion, [H₃O⁺], is 0.10 M

Aii. The concentration hydroxide ion, [OH⁻] is 1×10⁻¹³ M

Bi. The concentration of hydronium, ion [H₃O⁺], is 3.57×10⁻¹¹ M

Bii. The concentration hydroxide ion, [OH⁻] is 2.8×10¯⁴ M

A. How do i determine [H₃O⁺] and [OH⁻] of 0.10 M HCl?

i. The concentration of hydronium ion, [H₃O⁺] can be obtained as follow:

HCl(aq) + H₂O <=> H₃O⁺(aq) + Cl⁻(aq)

From the above equation,

1 mole of HCl contains 1 mole of H₃O⁺

Therefore,

0.10 M HCl will also contain 0.10 M H₃O⁺

Thus, the concentration of hydronium ion, [H₃O⁺] is 0.10 M

ii. The concentration of hydroxide ion, [OH⁻] can be obtained as follow:

Concentration of hydronium, ion [H₃O⁺] = 0.10 MConcentration hydroxide ion, [OH⁻] =?

[H₃O⁺] × [OH⁻] = 10¯¹⁴

0.10 × [OH⁻] = 10¯¹⁴

Divide both side by 3.02×10⁻¹⁰

[OH⁻] = 10¯¹⁴ / 0.10

[OH⁻] = 1×10⁻¹³ M

Thus, concentration of hydroxide ion, [OH⁻] is 1×10⁻¹³ M

B. How do i determine [H₃O⁺] and [OH⁻] for 1.4×10¯⁴ M Mg(OH)₂?

First, we shall obtain concentration hydroxide ion, [OH⁻]. Details below:

Mg(OH)₂(aq) <=> Mg²⁺(aq) + 2OH⁻(aq)

From the above equation,

1 mole of Mg(OH)₂ is contains 2 mole of OH⁻

Therefore,

1.4×10¯⁴ M Mg(OH)₂ will contain = 1.4×10¯⁴ × 2 = 2.8×10¯⁴ M OH⁻

Thus, concentration hydroxide ion, [OH⁻] is 2.8×10¯⁴ M

Now, we shall obtain the concentration of hydronium, ion [H₃O⁺]. Details below:

Concentration of hydroxide ion, [OH⁻] = 2.8×10¯⁴MConcentration of hydronium, ion [H₃O⁺] = ?

[H₃O⁺] × [OH⁻] = 10¯¹⁴

[H₃O⁺] × 2.8×10¯⁴ = 10¯¹⁴

Divide both side by 2.8×10¯⁴

[H₃O⁺] = 10¯¹⁴ / 2.8×10¯⁴

[H₃O⁺] = 3.57×10⁻¹¹ M

Thus, the concentration of hydronium, ion [H₃O⁺], is 3.57×10⁻¹¹ M

Learn more about hydroxide ion concentration, [OH⁻]:

#SPJ1