A student is given 3 beakers:

Beaker 1- 50.0 ml of a solution produced by dissolving 6.00 grams of a weak
monoprotic acid ,HX, in enough water to produce 1 liter of solution.
The empirical formula of HX is CH2O. The solution contains 3 drops
of phenolphthalein.

Beaker 2- A 0.07M solution of the salt NaX. It has a pH of 8.8

Beaker 3 – 50.0 ml of 0.250M KOH

The contents of beaker 3 is added drop-wise to beaker 1 until a pink color appears and remains for 30 seconds. This takes exactly 20.0 ml of the beaker 3 solution. Identify X and calculate the pH of beaker 1 after the addition of the 20 ml.

Beaker 2 solves for Ka.

...........X^- + HOH ==> HX + OH^-
initial...0.07...........0......0
change.....-y............y.....y
equil.....0.07-y.........y......y

Kb for X^- = (Kw/Ka for HX) = (y)(y)/(0.07-y)
Kw, of course, is 1E-14 and you solve for Ka for HX. y = OH and you can obtain that from the pH of 8.8.

Back to beaker #1.
6.00g/molar mass CH2O = 0.2M.
50 mL x 0.2M = 10 millimoles of HX
20 mL of 0.250 KOH (beaker 3) = 5 mmol.
Reaction of HX(beaker 1) and KOH (beaker 3).
.........HX + KOH ==> KX + H2O
initial..10.....0.......0....0
add............5.0...............
change...-5.0..-5.0....+5.0....+5.0
equil....5.0.....0.......5.0....5.0

Substitute from the above ICE chart into the Henderson-Hasselbalch equation and solve for pH.
pH = pKa + log [(base)/(acid)]
You know pKa from Ka you solved for from beaker 2.
Look up Ka in a table of Ka. I suspect that is formic acid and X^- is formate ion.
Post your work if you get stuck.

So, if the X suppose to be HCH2O?

how do find the pH of beaker 1 after the addition of the 20 mL

To identify X and calculate the pH of beaker 1 after the addition of the 20 ml, we need to understand the reaction that is taking place.

Based on the given information, we can determine the chemical equation for the reaction between the weak monoprotic acid (HX) and the strong base (KOH). The reaction can be represented as follows:

HX + KOH -> KX + H2O

From the equation, we see that HX reacts with KOH to form KX (the salt) and water. This means that X corresponds to the anion in the salt NaX, which is formed as a result of the reaction.

Given that NaX is a salt, it dissociates in water to form Na+ cations and X- anions. Therefore, the anion X is represented by X-.

To calculate the pH of beaker 1 after the addition of the 20 ml, we need to determine the final concentration of HX in the solution.

Initially, the solution contains 6.00 grams of HX in enough water to produce 1 liter of solution. Since the empirical formula of HX is CH2O, let's calculate the molar mass of CH2O:

Molar mass of C = 12.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of O = 16.00 g/mol

Molar mass of CH2O = (12.01 g/mol) + (2 * 1.01 g/mol) + (16.00 g/mol) = 30.03 g/mol

Now, we can convert the given mass of HX into moles:

Moles of HX = (6.00 g)/(30.03 g/mol) = 0.1999 mol (approximately 0.200 mol)

Since the solution is prepared in a 1 liter volume, the initial concentration of HX can be calculated as follows:

Initial concentration of HX = (0.200 mol)/(1.00 L) = 0.200 M

Upon adding 20.0 ml (or 0.020 L) of the 0.250 M KOH solution to the beaker 1, we can calculate the moles of KOH added:

Moles of KOH = (0.250 mol/L) * (0.020 L) = 0.005 mol

Since the reaction between HX and KOH occurs in a 1:1 ratio, the amount of HX reacted is equal to the moles of KOH added.

Moles of HX reacted = 0.005 mol

To determine the moles of HX remaining after the reaction, we subtract the moles reacted from the initial moles:

Moles of HX remaining = (0.200 mol) - (0.005 mol) = 0.195 mol

The final volume of the solution is the sum of the initial volume (50.0 ml) and the volume of KOH added (20.0 ml), which gives a total volume of (50.0 ml + 20.0 ml) = 70.0 ml = 0.070 L.

Finally, we can calculate the final concentration of HX:

Final concentration of HX = (0.195 mol)/(0.070 L) = 2.79 M (approximately)

Now, let's calculate the pH of beaker 1 after the addition of the 20 ml:

pH = -log10(H+ concentration)

Since the weak monoprotic acid HX has dissociated into one H+ ion and one X- ion:

[H+] = [X-] = 2.79 M

pH = -log10(2.79) = 0.555

Therefore, the pH of beaker 1 after the addition of the 20 ml is approximately 0.555.