Question

find pH

we ? for calculehen Calculated pH Measured pH 4.54 deionized water should be 7 40 mL DI water + 1.0 mL of 6.0 M HCl(aq) 40 mL DI water + 1.0 mL of 6.0 M NaOH(aq) 1.57 11.93 Compare the change in pH you observed when 1.0 mL of 6.0 M HCl is added to 40 mL of the buffer versus 40 mL of water. Is there a difference, and if so, why? Do the same for your observation of the change in pH of the buffer versus the water upon the addition of 1.0 mL of 6.0 M NaOH solution.

is this right im confusing myself a bit

Answer 1

pH of Water solution : pH = -log (H⁺)

1. deionized water : (H⁺) (OH^-) = 10^-14 ; so (H⁺) = 10^-7

   pH = -log (10^-7) = 7.00 m

2. 1.0 ml of 6.0 M HCl to 40 ml ,

HCl (aq) -------> H⁺ (aq) + Cl^- (aq)

deionized water : (H⁺) is negligible.

(H⁺) = 6.0* 1.0 ml /41 ml = 0.146 M

pH = -log ( 0.146) = 0.84

3. 1.0 ml of 6.0 M NaOH to 40 ml ,

NaOH (aq) -------> Na⁺ (aq) + OH^- (aq)

deionized water : (OH^-) is negligible.

((OH^-)) = 6.0* 1.0 ml /41 ml = 0.146 M

pH = 14 - pOH = 14 - ( -log ( 0.146)) = 13.16

pH for buffer solution :

Your volume of prepared solution of CH₃COONa and Molarity of CH₃COOH are not clear.

   pH = pKa + log ((A^-) /(HA))

pKa Acetic acid = 4.74

your data not clear there is guide how to calculate pH :

(a)   pH = pKa + log ((A^-) /(HA)) = 4.74 + log ((CH₃COONa) /(CH₃COOH))

(b) addition of HCl

reaction CH₃COO^- (aq) + H⁺ (aq) -----> CH₃COOH (aq) + H2O(l)

Calculate change in concentrations :

(CH₃COOH) = (40 ml* (CH₃COOH) M + 1ml *6.0 M )/ 41 ml =  

(CH₃COONa) = (40 ml* (CH₃COONa) - 1ml *6.0 M ) / 41 ml = 0.353 M

pH = 4.74 + log  ((CH₃COONa) /(CH₃COOH))

(c)

addition of NaOH

reaction CH₃COOH(aq) + OH^- (aq) -----> CH₃COO^- (aq) + H2O(l)

Calculate change in concentrations :

(CH₃COOH) = (40 ml* (CH₃COOH) M - 1ml *6.0 M )/ 41 ml =  

(CH₃COONa) = (40 ml* (CH₃COONa) + 1ml *6.0 M ) / 41 ml = 0.353 M

pH = 4.74 + log  ((CH₃COONa) /(CH₃COOH))