Find the pH of a 0.350 M aqueous benzoic acid solution. For benzoic add. Ka = 6.5 x 10^-5. a) 4.64
b) 4.19
c) 2.32
d) 11.68

correct option is (a)

The solution would be using this: C6H5COOH = H+ + C6H5COO Ka = 6.5 x 10^-5 = (H+)(C6H5COO-) over

(C6H5COOH)

Let X = moles per liter (H+) and also = moles per liter (C6H5COO-)

Ka = 6.5 x 10^-5 = (X)(X) over .350 molar = acid solution 6.5 x 10^-5 = X^2 over .350

X^2 = 6.5 x 10^-5 times .350 which = 2.275 x 10^-5

x = V2.275 x 10^-5

X = 1.5083 x 10^-5 moles per liter H+

pH = -log(H+) = -log 1.5083 x 10^-5 which

= 4.6215

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

explanation: