Question 7 7 pts A titration of a 108.0 mL aqueous solution of 0.13 MHI with...

Question

Question

Answer 1

Answer #1

HI ------------------------------------- NaOH

MA = 0.13M ----------------------- MB = 0.11M

VA = 108ml ------------------------VB = 111ml

M = MAVA-MBVB/VA + VB [ MAVA >MBVB]

= 0.13*108-0.11*111/(108+111)

= 1.83/219

= 0.008356

[H^+] = M = 0.008356M

PH = -log[H^+]

= -log0.008356

= 2

1.9>>>>answer

Answer 2

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