What is the pH when 0.40 g of NaOH is added to 50,0 mL of 0.10 M HCOOH? No volume change. (Ka for HCOOH - 1.8 x 10^-4.)
Convert 0.40 g NaOH to moles.
Convert 50.0 mL x 0.1 M HCOOH to moles.
See which is in excess. If NaOH, then calculate pOH and pH.
If HCOOH, it will be a buffer and you should use the Henderson-Hasselbalch equation.
To find the pH when NaOH is added to HCOOH, we need to consider the reaction that occurs between the two.
The balanced equation for the reaction between NaOH and HCOOH is:
HCOOH + NaOH -> HCOONa + H2O
From the balanced equation, we can see that 1 mole of HCOOH reacts with 1 mole of NaOH to produce 1 mole of HCOONa.
Given that the molar mass of NaOH is 39.997 g/mol and the molar mass of HCOOH is 46.0254 g/mol, we can calculate the number of moles of NaOH added:
0.40 g NaOH * (1 mol NaOH / 39.997 g NaOH) = 0.010 mol NaOH
Since there is no change in volume, the initial volume of HCOOH is still 50.0 mL or 0.050 L.
The initial number of moles of HCOOH can be calculated using the concentration:
mol HCOOH = concentration HCOOH * volume HCOOH
= 0.10 M * 0.050 L
= 0.005 mol HCOOH
By comparing the stoichiometric coefficients, we can see that 0.005 mol of HCOOH reacts with 0.005 mol of NaOH. Since NaOH is in excess, all the HCOOH will react, and we will have 0 moles of HCOOH remaining.
To calculate the concentration of HCOO− (the conjugate base of HCOOH), we divide the number of moles of HCOO− by the final volume:
concentration HCOO− = (0.005 mol HCOO−) / 0.050 L
= 0.10 M
Now, we can use the given Ka value of HCOOH (1.8 x 10^-4) to calculate the concentration of H3O+ (the hydronium ion) using the following equation:
Ka = ([H3O+][HCOO−]) / [HCOOH]
Since the concentration of HCOOH is now 0, we can simplify the equation to:
Ka = [H3O+][HCOO−]
Plugging in the given Ka value, we have:
1.8 x 10^-4 = [H3O+][0.10 M]
Rearranging the equation:
[H3O+] = (1.8 x 10^-4) / (0.10 M)
= 1.8 x 10^-3 M
Finally, we can calculate the pH using the formula:
pH = -log[H3O+]
pH = -log(1.8 x 10^-3)
pH = 2.74
Therefore, when 0.40 g of NaOH is added to 50,0 mL of 0.10 M HCOOH (with no volume change), the pH of the resulting solution is approximately 2.74.