A solution of volume 80.0 mL contains 16.0 mmol HCHO2 and 9.00 mmol NaCHO2.
If 1.00 ml of 12 M HCl is added to this, what will be the resulting pH?
(the only problem im having for this question is how to approach it, i know that by adding HCl the concentration of the CB would decrease and the concentration of the acid would increase, but how do you factor in the extra 1ml added because of HCL? )
To approach this problem, we can start by calculating the initial concentrations of the acid (HCHO2) and conjugate base (NaCHO2) in the solution before adding HCl.
Given:
Volume of solution = 80.0 mL
HCHO2 concentration = 16.0 mmol / 80.0 mL = 0.20 M
NaCHO2 concentration = 9.00 mmol / 80.0 mL = 0.113 M
Now, let's consider the reaction that occurs when HCl is added to the solution:
HCl(aq) + HCHO2(aq) -> H3O+(aq) + CHO2-(aq)
Given that 1.00 mL of 12 M HCl is added, the moles of HCl added can be calculated as follows:
Moles of HCl = Volume of HCl (L) x Concentration of HCl (mol/L)
= 0.001 L x 12 mol/L
= 0.012 mol
Since HCHO2 and HCl are in a 1:1 stoichiometric ratio, the moles of HCHO2 remaining will be:
Moles of HCHO2 remaining = Moles of HCHO2 initially - Moles of HCl added
= 16.0 mmol - 12.0 mmol
= 4.00 mmol
The total volume of the solution after adding HCl would be 80.0 mL + 1.00 mL = 81.0 mL.
Now, let's calculate the concentrations of HCHO2 and NaCHO2 after the addition of HCl:
HCHO2 concentration = (4.00 mmol) / (81.0 mL) = 0.049 M
NaCHO2 concentration = 9.00 mmol / 81.0 mL = 0.111 M
To determine the resulting pH, we need to compare the concentrations of the weak acid and its conjugate base. Since the HCHO2 concentration is greater than the NaCHO2 concentration, the resulting solution will be acidic.
To calculate the pH, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ( [A-] / [HA] )
The pKa value for formic acid (HCHO2) is 3.75.
Using the equation, the pH can be calculated as:
pH = 3.75 + log ( 0.111 / 0.049 )
Calculating this expression gives us a pH of approximately 3.97.
Therefore, the resulting pH after adding 1.00 mL of 12 M HCl to the solution will be approximately 3.97.
To approach this problem, we need to consider the reaction that occurs between HCHO2 (the weak acid) and HCl (the strong acid). HCHO2 will be protonated by HCl, forming the conjugate acid of HCHO2, and chloride ions (Cl-) will be formed as a byproduct. The resulting solution will contain the conjugate acid, HCHO2, and its conjugate base, NaCHO2.
Let's break down the steps to solve this problem:
1. Calculate the initial concentrations of HCHO2 and NaCHO2 before adding HCl:
- HCHO2: 16.0 mmol / 80.0 mL = 0.200 mol/L = 0.200 M
- NaCHO2: 9.00 mmol / 80.0 mL = 0.1125 mol/L = 0.1125 M
2. Determine the volume of the resulting solution after adding 1.00 mL of 12 M HCl:
- The total volume of the solution will be 80.0 mL + 1.00 mL = 81.0 mL
3. Calculate the moles of HCl added:
- Moles of HCl = volume (L) x concentration (mol/L)
- Moles of HCl = 0.001 L x 12.0 mol/L = 0.012 mol
4. Determine the final concentrations of HCHO2, NaCHO2, and HCl after the reaction:
- HCHO2: 16.0 mmol - 0.012 mol / 81.0 mL = 0.188 M
- NaCHO2 remains unchanged because it does not react with HCl.
- HCl: 0.012 mol / 81.0 mL = 0.148 M
5. Calculate the concentration of the conjugate base, CHO2- (NaCHO2 dissociates to produce CHO2- ions):
- CHO2-: 9.00 mmol / 81.0 mL = 0.111 M
6. Determine the pKa of HCHO2:
- The pKa of HCHO2 can be found in a table, or you can calculate it using Ka = [H+][CHO2-] / [HCHO2].
- The pKa of HCHO2 is approximately 3.75.
7. Use the Henderson-Hasselbalch equation to calculate the pH of the resulting solution:
- pH = pKa + log (base/acid)
- pH = 3.75 + log (0.111 / 0.188)
By following these steps, you can determine the resulting pH of the solution after adding 1.00 mL of 12 M HCl.