Question

Consider the titration of 50.0 ml of 0.314 m weak base b ( kb =7.5*; 10 ^ - 6 ) with 0.340 m hn*o_{3} what is the ph of the solution after the addition of 90.0 ml of hn*o_{3} ?

Answer 1

The pH of the solution after the addition of 90.0 mL of 0.340 M HNO3 to 50.0 mL of 0.314 M weak base B is approximately 0.94.

Step-by-step explanation:

To calculate the pH of a solution after the titration of a weak base with a strong acid, you need to consider the amounts of substance involved and the dissociation of the weak base. In this case, we have a titration of 50.0 mL of 0.314 M weak base B with 0.340 M HNO3.

We start by calculating the moles of the weak base and the moles of the strong acid added:

• Moles of weak base B = 0.0500 L × 0.314 M = 0.0157 moles
• Moles of HNO3 added = 0.0900 L × 0.340 M = 0.0306 moles

Since HNO3 is a strong acid, it will react completely with the weak base, forming the conjugate acid of the weak base and water. After the reaction, there will be an excess of HNO3 since moles of HNO3 > moles of the weak base B:

Excess moles of HNO3 = 0.0306 moles - 0.0157 moles = 0.0149 moles

The concentration of the excess HNO3 in the total volume of the solution is calculated as:

Concentration of HNO3 = 0.0149 moles / (0.0500 L + 0.0900 L) = 0.114 M

Since we are dealing with a strong acid, the pH can be found by taking the negative logarithm of the hydronium ion concentration, which is equal to the concentration of the acid in solution:

pH = -log([H+]) = -log(0.114) ≈ 0.94