PH chemistry question help!?
Could someone help with this question? And explain why
A known volume )1mL) of a 0.100M NaOH solution was diluted to a total volume of 10.0mL (Solution A). Next 1.00mL of this solution was diluted to a total volume of 50.mL (Solution B) What is the pH of solution B?
A known volume )1mL) of a 0.100M NaOH solution was diluted to a total volume of 10.0mL (Solution A). Next 1.00mL of this solution was diluted to a total volume of 50.mL (Solution B) What is the pH of solution B?
Best Answer - Chosen by Voters
Using M1V1 = M2V2,
0.1 x 1 = M2 x 10
M2 = 0.01M (solution A)
Repeat the same process,
0.01 x 1 = M3 x 50
M3 = 0.0002 M (solution B)
Hence , [OH-] = 0.0002M
pOH = -log [OH-]
= 3.70
pH = 14 - 3.70 = 10.30
0.1 x 1 = M2 x 10
M2 = 0.01M (solution A)
Repeat the same process,
0.01 x 1 = M3 x 50
M3 = 0.0002 M (solution B)
Hence , [OH-] = 0.0002M
pOH = -log [OH-]
= 3.70
pH = 14 - 3.70 = 10.30
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