Abeaker with 1.80×102 ml of an acetic acid buffer with a ph of 5.000 is sitting on a benchtop. the total molarity of acid and conjugate base in this buffer is 0.100 m. a student adds 5.10 ml of a 0.440 m hcl solution to the beaker. how much will the ph change? the pka of acetic acid is 4.740.

The pH will become a little more acid, by changing 0.225 units.

Explanation:

Step 1: Data given

Volume of acetic acid buffer = 1.80 * 10² mL

pH of the buffer = 5.00

Molarity of acid and conjugate base = 0.1.00 M

A student adds 5.10 mL ( = 0.0051 L) of a 0.440 M HCl solution

The pKa of acetic acid is 4.74

Step 2: The Henderson-Hasselbalch equation

pH = pKa + log ([conjugate base]/[weak acid])

with acetic acid (CH3COOH) as the weak acid and CH3COO- as it's conjugate base. The pH =5 and the pKa of acetic acid is 4.74

5 = 4.74 +  log ([conjugate base]/[weak acid])

log ([CH3COO-]/[(CH3COOH]) = 0.26

[CH3COO-]/[(CH3COOH] = 10^0.26 = 1.82

This means the buffer contains 1.82 times more conjugate base (CH3COO-) than weak acid (CH3COOH)

Since both the conjugate base and the weak acid, share the same volume we can say:

(Number of moles of CH3COO-) = 1.82* (Number of moles of CH3COOH)

Step 3: Calculate number of moles of acid and conjugate base

As said, the total molarity of acid and conjugate base is 0.100 M

[CH3COOH] + [CH3COO-] = 0.100 M

Since molarity = Number of moles/ volume  (mol/L), We can write this as:

(n(CH3COOH)/180 * 10^-3 L ) + (n(CH3COO-)/180 * 10^-3 L ) = 0.100 moles / L

OR

n(CH3COOH) + n(CH3COO-) = 0.018

1.82*n(CH3COOH) + n(CH3COOH) = 0.018

n(CH3COOH) = 0.00638 moles

n(CH3COO-) = 1.82 * 0.00638 moles = 0.0116 moles

Step 4: Adding  5.10 mL of a 0.440 M HCl solution:

When adding HCl, the HCl will  react with the acetate anions to form acetic acid and chloride anions ( Cl-):

HCl(aq) + CH3COO- (aq) → CH3COOH(aq) + Cl-(aq)

⇒For 1 mole of HCl consumed, we need 1 mole of CH3COO-, to produce 1 mole of CH3COOH and 1 mole of Cl-

Step 5: Calculate moles of HCl

Number of moles = Molarity * Volume

Number of moles HCl = 0.440 M * 5.10 *10^-3 L =  0.002244 moles

Step 6: The limiting reactant

HCl is the limiting reactant and will be completely consumed by the reaction. 0.002244 moles is consumed. There will remain 0 moles.

Step 7: Calculate remain moles

For CH3COO- , there is also 0.002244 moles consumed.

There will remain 0.0116 moles - 0.002244 moles = 0.009356 moles

For CH3COOH, there will be produced 0.002244 moles. There will be in total: 0.00638 moles + 0.002244 moles = 0.008624 moles

The total volume of the solution is: 180 mL + 5.10 mL = 185.1 mL = 0.1851 L

Step 8: Calculate the concentrations

Concentration = Number of moles / volume

The concentrations of acetic acid and acetate ions will be:

[CH3COOH] =  0.008624 moles / 0.1851 L

[CH3COOH] = 0.0466 M

[CH3COO-] = 0.009356 moles / 0.1851 L

[CH3COO-] = 0.0505 M

Step 9: Calculate the new pH

pH = 4.74 + log(0.0505 M/ 0.0466 M)

pH = 4.775

ΔpH = 5 - 4.775 = 0.225

The pH will become a little more acid, by changing 0.225 units.

the dependant variable is spilled water

the independent is the heat