Question

Consider the titration of a 21.0mL sample of 0.110 M HC2H3O2 with 0.125 M NaOH. (The value of Ka for HC2H3O2 is 1.8×10−5.) Part A: Determine the initial pH. Part C: Determine the pH at 5.0 mL of added base.

Answer 1

NaClO₂ + 1₂0 - HGM, ² + Naon QIXO11 0.125xo Initially only weak acid is present Sit dissociates to give crtion, we can use the following formula. So [+] = Kane [MtJ = V1-8x105*0.11 In+] = 50. 198x10-5 [ht] => 7988106 [+] = 1.400103 ph=-14+3 =- logic4 x16 pa = 3-log 1.4 pH = 3-0.14 on = 2.86

Post > N4 с. Изо, тн, о ии, O, TN4 ои gloooll 5x00125 2-31 0.625 22310625 Acidic Buffer 0.625 5- در ۱۰۶ - ا ph= pka + log (salt) ph - 4.75+ logotes pk =4.75+ logo-37 P = 4.75 -0.43 Tph = 4.32 qka = -dog ka pka = - dog 18x10 pla = 5-dog 1-8 . pka = 5-0.25 plez 4.95
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