Question

A solution is prepared by dissolving 0.23 mol of nitrous acid and 0.27 mol of sodium nitrite in water sufficient to yield 1.00 L of solution. The addition of 0.05 mol of HCl to this buffer solution causes the pH to drop slightly. The pH does not decrease drastically because the HCl reacts with the __________ present in the buffer solution. The Ka of nitrous acid is 1.36 × 10-3.A) H2OB) H3O+C) nitrite ionD) nitrous acidE) This is a buffer solution: the pH does not change upon addition of acid or base.

Answer 1

Complete Question

A solution is prepared by dissolving 0.23 mol of nitrous acid and 0.27 mol of sodium nitrite in water sufficient to yield 1.00 L of solution. The addition of 0.05 mol of HCl to this buffer solution causes the pH to drop slightly. The pH does not decrease drastically because the HCl reacts with the __________ present in the buffer solution. The Ka of nitrous acid is 1.36 × 10-3.

A) H2O

B) H3O+

C) nitrite ion

D) nitrous acid

E) This is a buffer solution: the pH does not change upon addition of acid or base.

Answer:

The correct option is  C

Explanation:

From the question we are told that

           The number of moles of nitrous acid is  $n = 0.23 \ mols$

          The number of moles of sodium nitrite is $m = 0.27 \ moles$

          The volume of the solution is  $V = 1.00\ L$

           The number of mole added is  $z = 0.05 \ moles$

             The Ka for nitrous acid is  $Ka = 1.36 *10^{3}$

Now looking at the question  we can see that the solution is  made up of

     A weak acid , and its salt and a strong base and this would imply that the buffer is an acidic buffer what this mean is that  the  addition of small amount of  acid to the solution would pH

   Sine Nitrite  ion is the basic component of the buffer the it then means that the HCl would react with the Nitrite  

Answer 2

A buffer contains 0.23 mol of nitrous acid and 0.27 mol of sodium nitrite in 1.00 L of solution. Upon addition of 0.05 moles of HCl, the pH does not decrease drastically because the HCl reacts with the C) nitrite ion present in the buffer solution.

A solution is prepared by dissolving 0.23 mol of nitrous acid (weak acid) and 0.27 mol of sodium nitrite (conjugate base) in water sufficient to yield 1.00 L of solution. The corresponding concentrations are:

$[HNO_2] = \frac{0.23mol}{1.00L} = 0.23 M\\[NaNO_2] = \frac{0.27mol}{1.00L} = 0.27 M$

We can calculate the pH of this buffer using Henderson-Hasslebach's equation.

$pH = pKa + log\frac{[NaNO_2]}{[HNO_2]} = -log(1.36 \times 10^{-3} ) + log (0.27/0.23) = 2.94$

When 0.05 moles of HCl are added, they react with the conjugate base. The net ionic equation is:

                   HCl  +  NO₂⁻ ⇒  HNO₂  +  Cl⁻

Initial           0.05     0.27        0.23        0

Reaction    -0.05    -0.05      +0.05     +0.05

Final              0        0.22        0.28       0.05

We can calculate the pH of the new solution using Henderson-Hasslebach's equation.

$pH = pKa + log\frac{[NaNO_2]}{[HNO_2]} = -log(1.36 \times 10^{-3} ) + log (0.22/0.28) = 2.76$

Upon the addition of HCl, the pH went from 2.94 to 2.76. It didn't decrease drastically because HCl reacted with NO₂⁻.

A buffer contains 0.23 mol of nitrous acid and 0.27 mol of sodium nitrite in 1.00 L of solution. Upon addition of 0.05 moles of HCl, the pH does not decrease drastically because the HCl reacts with the C) nitrite ion present in the buffer solution.

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