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.

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )