What is the ph of a solution prepared by mixing 25.00 ml of 0.10 m ch3co2h with 25.00 ml of 0.010 m ch3co2na? assume that the volume of the solutions are additive and that k a = 1.8 × 10-5 for ch3co2h.

pH = 3.74

Explanation:

Given:

Initial volume of CH3COOH, V1 = 25.00 ml

Initial concentration of CH3COOH, M1 = 0.10 M

Initial volume of CH3COONa, V1 = 25.00 ml

Initial concentration of CH3COONa, M2 = 0.010 M

Ka (CH3COOH) = 1.8*10^-5

To determine:

pH of the solution

Calculation:

The given solution of CH3COOH/CH3COONa is in fact a buffer whose pH is given by the Henderson-Hasselbalch equation where:

where A- = concentration of conjugate base = [CH3COONa]

HA = weak acid = [CH3COOH]

Step 1: Calculate the final concentration of CH3COONa

V1 = 25.00 ml

V(final) = Total volume = 25.00 + 25.00 = 50.00 ml

M1 = 0.010 M

Step 2: Calculate the final concentration of CH3COOH

V1 = 25.00 ml

V(final) = Total volume = 25.00 + 25.00 = 50.00 ml

M1 = 0.10 M

Step 3: Calculate the pH

Based on equation (1)

pKa = -log Ka = -log(1.8*10^-5) = 4.74

pH = 3.74

the formula would be mgso4

mgso[subscript]4

answer: your mm

explanation: