How many milliliters of 0.246 M HNO3 should be added to 213 mL of 0.00666M 2,2'-bipyridine to give a pH of 4.19?
To solve this problem, we need to use the concept of acid-base reactions and the equation for pH. Here's how you can approach it:
Step 1: Write the equation for the acid-base reaction.
In this case, we have HNO3 (nitric acid) as the acid and 2,2'-bipyridine as the base. The reaction can be written as follows:
HNO3 + 2,2'-bipyridine -> H2O + [2,2'-bipyridine-H]+ + [NO3]-
Step 2: Calculate the concentration of [2,2'-bipyridine-H] ions.
Since we know the pH, we can calculate the concentration of [2,2'-bipyridine-H] using the equation for pH:
pH = -log[H+]
Rearranging the equation:
[H+] = 10^(-pH)
[H+] = 10^(-4.19)
Step 3: Calculate the required amount of HNO3.
Now, we can use the equation for the acid-base reaction to determine the moles of acid required to neutralize the [2,2'-bipyridine-H] ions. Since we know the concentration of 2,2'-bipyridine and the concentration of [2,2'-bipyridine-H], we can use the equation:
moles of acid = moles of base
moles of HNO3 = (concentration of [2,2'-bipyridine-H] ions) * (volume of 2,2'-bipyridine)
Step 4: Convert the moles of HNO3 to milliliters.
Since we have the concentration of HNO3, we can convert the moles of HNO3 to milliliters using the equation:
moles of HNO3 = (concentration of HNO3) * (volume of HNO3 in milliliters)
Solving this equation will give us the volume of HNO3 required to neutralize the [2,2'-bipyridine-H] ions.
By following these steps and plugging in the given values, you should be able to find the answer.