Question

Calculate the ph of a 0.51 m ch3coona solution. (ka for acetic acid = 1.8x10^-5).

Answer 1

option b) 294

hope you have got answer

Answer 2

Given:

A 0.51 m ch3coona solution. (ka for acetic acid = 1.8x10^-5).

To Find:

Calculate the ph of a 0.51 m ch3coona solution. (ka for acetic acid = 1.8x10^-5).

Solution:

To find the ph of a 0.51 m ch3coona solution we will follow the following steps:

As we know,

ch3coona is a salt of a weak acid ch3cooh and a strong base NaOH.

The formula for finding the PH of a weak acid and strong base is:

$PH = 7 + \frac{1}{2}pka \: + \frac{1}{2} log \: c$

Ka is the ionisation constant for acetic acid and c is the concentration of ch3coona which is equal to 0.51m.

Ch3coona dissociate into CH3coo- and Na+.

Now,

Putting the values of Ka and concentration we get,

$PH = 7 + \frac{1}{2}p(1.8 \: \times \: 10^-5) \: + \frac{1}{2} log \: 0.51$

log 0.51 = -0.29

and PKA of ka = (1.8 × 10-5) = -log(Ka) -log(1.8)+-(-5log10) = -0.255+5 = 4.74

$PH = 7 + \frac{1}{2}(4.74) \: + \frac{1}{2} (-0.29)$

PH = 7+2.37-0.145

PH = 7+2.22 = 9.22

Henceforth, the ph of a 0.51 m ch3coona solution is 9.22.