The pH of the nitrite buffer has changed from 3.00 to 2.80 by adding 11.0mL of 12.0M HCL. How many grams of solid sodium Hydroxide would the chemist add to the buffer to change the pH from 3.00 to 3.80?
If I use a pKa of 3.29 the addition of 11.0 mL of 12.0M HCl will lower the pH to 2.71 which means you made a typo or we aren't using the same pKa.
What pKa do you have for HNO2?
To solve this problem, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the concentrations of the acid and conjugate base components of the buffer.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
- pH is the desired pH of the buffer (3.80 in this case)
- pKa is the dissociation constant of the acid (1.94 for nitrous acid, HNO2)
- [A-] is the concentration of the conjugate base (nitrite, NO2-) in the buffer solution
- [HA] is the concentration of the acid (nitrous acid, HNO2) in the buffer solution
In this question, the initial pH is 3.00, and we want to change it to 3.80. Therefore, we can set up the equation:
3.80 = pKa + log([A-]/[HA])
Since the buffer is initially at pH 3.00, we can assume that [A-] and [HA] are equal. Let's call their concentration x.
3.80 = 1.94 + log(x/x)
Now, let's solve for x (the concentration of [A-] and [HA]).
3.80 - 1.94 = log(x/x)
1.86 = log(x/x)
To convert the logarithmic equation to exponential form, we rewrite it as follows:
10^1.86 = x/x
x = 10^1.86
Now, we have the concentration of [A-] and [HA].
Next, we need to find the number of moles of nitrous acid in the solution. We can use the formula:
moles = concentration (mol/L) x volume (L)
The given volume of HCl added is 11.0 mL, which we need to convert to liters:
11.0 mL = 11.0 x 10^(-3) L
Now, we can calculate the moles of nitrous acid:
moles HNO2 = [HA] x volume = x x 11.0 x 10^(-3)
Since the concentration of the HCl is 12.0 M, we can use the relationship:
moles HCl = concentration (mol/L) x volume (L)
moles HCl = 12.0 M x 11.0 x 10^(-3) L
Now, we need to find the reaction stoichiometry between HCl and HNO2 to determine the number of moles of sodium hydroxide (NaOH) required to neutralize the excess HCl.
The balanced equation is:
HNO2 + NaOH → NaNO2 + H2O
From the balanced equation, we can see that 1 mole of HCl reacts with 1 mole of HNO2.
Therefore, the moles of HNO2 reacted with HCl is equal to the moles of HCl.
Now, we have the moles of NaOH required to neutralize the HCl, which is equal to the moles of HNO2 reacted with HCl.
Finally, we can calculate the mass of NaOH using the formula:
mass = moles x molar mass
The molar mass of NaOH is 40.0 g/mol.
So, the mass of solid sodium hydroxide that the chemist needs to add to the buffer to change the pH from 3.00 to 3.80 is:
mass of NaOH = moles x molar mass = moles x 40.0 g/mol
Note: Substitute the calculated values for moles and solve to find the final answer.