Robin the Research scientist prepares 100 mLs of a solution of 100 mM of Tris HCl. What is the pH? and part two for this question
Calculate the pH when the scientist adds 100 ul of 50% NaOH
I am confused since we do not know the conjugate base concentration so how to solve for the pH
isn't the conjugate base H2O?
i guess so? i am not to sure
To calculate the pH of a solution, you need to know the concentration of hydrogen ions (H+). In this case, Tris HCl is a weak acid, which means it partially dissociates in water to produce H+ ions.
For the first part of the question, the concentration of Tris HCl is given as 100 mM. To find the pH, you can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this equation, pKa is the dissociation constant of the weak acid, [A-] is the concentration of the conjugate base (Tris-), and [HA] is the concentration of the weak acid (Tris HCl).
Since Tris HCl is a weak acid, it dissociates as follows:
Tris HCl ⇌ Tris- + H+
The pKa value for Tris HCl is around 7.8. Since the pH of the solution will be close to the pKa value, you can approximate it to 7.8.
Using the equation, let's assume that after the partial dissociation, the concentration of Tris- ([A-]) is equal to the concentration of Tris HCl ([HA]). So, the equation becomes:
pH = pKa + log([A-]/[HA])
pH = 7.8 + log(100 mM/100 mM) = 7.8 + log(1) = 7.8 + 0 = 7.8
Therefore, the pH of the 100 mM Tris HCl solution is 7.8.
Now, moving on to the second part of the question, where 100 μl (microliters) of 50% NaOH is added to the solution. NaOH is a strong base and completely ionizes in water to produce OH- ions.
To calculate the pH after adding NaOH, you need to consider the reaction between H+ and OH-:
H+ + OH- ⇌ H2O
Since the concentration of NaOH is not mentioned, we need to calculate the amount of OH- ions added when 100 μl of 50% NaOH is added to the 100 mL Tris HCl solution.
50% NaOH means that 50 grams of NaOH is dissolved in 100 ml of solution.
To calculate the moles of NaOH, you need to know the molar mass of NaOH, which is 40 g/mol (Na = 23 g/mol, O = 16 g/mol, H = 1 g/mol).
Moles of NaOH = (Mass of NaOH / Molar Mass of NaOH)
= (50 g / 40 g/mol)
= 1.25 mol
Since NaOH is a strong base, it dissociates to produce one mole of OH- ions per mole of NaOH:
Moles of OH- in 100 μl of 50% NaOH = Moles of NaOH = 1.25 mol
Now, to find the new concentration of OH- in the solution, you need to convert the volume to liters:
Volume of NaOH in liters = 100 μl * (1 ml / 1000 μl) * (1 L / 1000 ml)
= 0.0001 L
Concentration of OH- ions = Moles of OH- / Volume of solution in liters
= 1.25 mol / 0.0001 L
= 12,500 M
Using the equation for pH, which is:
pH = -log[H+]
To calculate the concentration of H+ ions, you need to consider that OH- and H+ ions react in a 1:1 ratio:
[H+] = [OH-] = 12,500 M
pH = -log(12,500) = -4.9
Therefore, the pH after adding 100 μl of 50% NaOH to the 100 mM Tris HCl solution is approximately -4.9.
Note: The negative pH value indicates that the solution is highly basic.