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when a 16.6 ml sample of a 0.407 m aqueous nitrous acid solution is titrated with a 0.439 m aqueous sodium hydroxide solution, what is the ph after 23.1 ml of sodium hydroxide have been added?

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Answer:

Step-by-step explanation:

The balanced chemical equation for the reaction between nitrous acid (HNO2) and sodium hydroxide (NaOH) is:

HNO2 + NaOH → NaNO2 + H2O

From the equation, we can see that 1 mole of NaOH reacts with 1 mole of HNO2. Therefore, the number of moles of NaOH added can be calculated as:

moles of NaOH = molarity x volume = 0.439 mol/L x 0.0231 L = 0.01013 mol

Since the balanced equation tells us that 1 mole of NaOH reacts with 1 mole of HNO2, the number of moles of HNO2 that reacted can also be calculated as 0.01013 mol.

The initial number of moles of HNO2 can be calculated from the concentration and volume of the solution:

moles of HNO2 = molarity x volume = 0.407 mol/L x 0.0166 L = 0.00675 mol

Therefore, the remaining number of moles of HNO2 after the reaction is:

remaining moles of HNO2 = initial moles of HNO2 - moles of NaOH used

remaining moles of HNO2 = 0.00675 mol - 0.01013 mol = -0.00338 mol

Since we cannot have a negative number of moles, this means that all the HNO2 has been consumed and there is an excess of NaOH. The excess NaOH will react with water to produce hydroxide ions (OH-), so we can calculate the concentration of OH- ions in the solution:

moles of OH- = moles of NaOH used = 0.01013 mol

volume of solution = initial volume of HNO2 solution + volume of NaOH solution = 0.0166 L + 0.0231 L = 0.0397 L

concentration of OH- ions = moles of OH- / volume of solution = 0.01013 mol / 0.0397 L = 0.255 M

Since we have a strong base (OH-) in solution, we can assume that the pH is basic. We can calculate the pOH of the solution as:

pOH = -log[OH-] = -log(0.255) = 0.594

The pH can be calculated from the pOH:

pH + pOH = 14

pH = 14 - pOH = 14 - 0.594 = 13.406

Therefore, the pH of the solution after 23.1 mL of NaOH have been added is approximately 13.4.

User Pavel Timoshenko
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