A25.0 ml sample of 0.150 m benzoic acid is titrated with a 0.150 m naoh solution. what is the ph at the equivalence point? the ka of benzoic acid is 4.50 × 10-4.

Solution:

At the equivalence point, moles NaOH = moles benzoic acid  

HA + NaOH ==> NaA + H2O where HA is benzoic acid  

At the equivalence point, all the benzoic acid ==> sodium benzoate  

A^- + H2O ==> HA + OH- (again, A^- is the benzoate anion and HA is the weak acid benzoic acid)  

Kb for benzoate = 1x10^-14/4.5x10^-4 = 2.22x10^-11  

Kb = 2.22x10^-11 = [HA][OH-][A^-] = (x)(x)/0.150  

x^2 = 3.33x10^-12  

x = 1.8x10^-6 = [OH-]  

pOH = -log [OH-] = 5.74  

pH = 14 - pOH = 8.26  

pH = 8.11

Explanation:

Step 1: Data given

Volume of a 0.150 M benzoic acid sample = 25.0 mL

Molarity of NaOH solution = 0.150 M

Ka of the benzoic acid = 4.50 * 10^-4

Step 2: The balanced equation

C7H6O2 + NaOH → C7H5O2Na  + H2O

Step 3: calculate moles

At the equivalence point we have the same amount of of acid and base

Both benzoic acid and NaOH will completely be consumed.

Moles benzoic acid = molarity *volume

Moles benzoic acid = 0.150 M * 0.025 L

Moles benzoic acid = 0.00375 moles

We we'll need 0.00375 moles NaOH to reach the equivalence point

Volume NaOH = 0.00375 moles / 0.150 M

Volume = 0.025 L = 25 mL

Total volume = 25 + 25 = 50 mL

Step 4: Calculate moles of C7H5O2Na

For 1 mol benzoic acid we need 1 mol NaOH to produce 1 mol C7H5O2Na

For 0.00375 moles benzoic acid we'll have 0.00375 moles C7H5O2Na

Step 5: The equation

C7H5O2- + H2O → C7H6O2 + OH-

Step 6: The initial concentration

[C7H5O2-] = 0.00375 moles / 0.050 L

[C7H5O2-] = 0.075 M

[C7H6O2] = 0M

[OH-] = 0M

Step 7: The concentration at the equilibrium

[C7H5O2-] = 0.075 - X M

[C7H6O2] = XM

[OH-] = XM

Step 8: Calculate Kb

Ka * Kb = Kw

Kb = Kw / Ka

Kb = 10^-14 / (4.50 *10^-4)

Kb = 2.22 *10^-11

Kb =  [C7H6O2][OH-]/[C7H5O2-]

2.22 * 10^-11 = X²/ 0.075 - X

2.22 * 10^-11 = X²/ 0.075

X² = 2.22 *10^-11 * 0.075

X² = 1.665 *10^-12

X = 1.29 * 10^-6 = [OH-]

Step 9: Calculate pOH

pOH = -log[OH-]

pOH = -log(1.29 * 10^-6 )

pOH = 5.89

Step 10: Calculate pH

pH = 14 - 5.89

pH = 8.11

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.