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What is the ph of a buffer made by combining 45. 0 ml of 0. 150 m nitrous acid and 20. 0 ml of 0. 175 m sodium nitrite? the ka of nitrous acid, hno2, is 4. 5×10^–4?

Sagot :

The pH of the buffer made by combining Nitrous acid and Sodium Nitrite will be 3.06.

pH can be defined as the medium of expressing the acidity or alkalinity of any solution. It is the potential of hydrogen atoms.

Buffers are the solutions that resist any change in the pH when an acid or base is added to them.

According to the question,

0.150 M nitrous acid (HNO₂) = 45 ml

0.175 M sodium nitrate (NO₂⁻)= 20 ml

Kₐ of nitrous acid (HNO₂) = 4.5 × 10⁻⁴

Total volume of buffer = 45 ml + 20 ml = 65 ml

Concentration of HNO₂ = (45) (0.150) / 65 = 0.103 M

Concentration of NO₂⁻ = (20) (0.175) / 65 = 0.054 M

                     HNO₂  ⇒ H⁺ + NO₂⁻

Initial :          0.103        0     0.054

Change:         -x             +x       +x

Equilibrium: 0.103-x        x       0.054+x

Kₐ = x(0.054+x) / 0.103-x = 0.054x / 0.103 = 0.524x

Also, Kₐ = 4.5 × 10⁻⁴

Therefore, 4.5 × 10⁻⁴ = 0.524x

                              x = 8.58 × 10⁻⁴ M

H⁺ = x = 8.58 × 10⁻⁴ M = 8.6 × 10⁻⁴ M

Therefore, pH = -log ( 8.6 × 10⁻⁴ M)

                 pH = 3.06

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