I am including all of the items because this is from one problem, unfortunately.
ID: 787486 • Letter: I
Question
I am including all of the items because this is from one problem, unfortunately. I would appreciate any help I can possibly get! Thank you!Part E Calculate [OH?] for 1.60mL of 0.180M NaOH diluted to 1.50L . Express your answer using three significant figures.
Part F Calculate pH for 1.60mL of 0.180M NaOH diluted to 1.50L . Express your answer using three decimal places.
Part G Calculate [OH?] for a solution formed by adding 5.30mL of 0.110M KOH to 20.0mL of 9.2 I am including all of the items because this is from one problem, unfortunately. I would appreciate any help I can possibly get! Thank you!
Part E Calculate [OH?] for 1.60mL of 0.180M NaOH diluted to 1.50L . Express your answer using three significant figures.
I am including all of the items because this is from one problem, unfortunately. I would appreciate any help I can possibly get! Thank you!
Part E Calculate [OH?] for 1.60mL of 0.180M NaOH diluted to 1.50L . Express your answer using three significant figures.
Part F Calculate pH for 1.60mL of 0.180M NaOH diluted to 1.50L . Express your answer using three decimal places.
Part F Calculate pH for 1.60mL of 0.180M NaOH diluted to 1.50L . Express your answer using three decimal places.
Part G Calculate [OH?] for a solution formed by adding 5.30mL of 0.110M KOH to 20.0mL of 9.2 Part G Calculate [OH?] for a solution formed by adding 5.30mL of 0.110M KOH to 20.0mL of 9.2
Explanation / Answer
E) moles of OH- = ( 0.0016 x 0.18) = 0.000288 + 10^-7 ( from neutral water)
now [OH-] = ( 0.0002887/1.5) = 0.000192 = 1.92 x10^-4
F) pOH = -log [OH-] = -log ( 1.9246 x10^-4) = 3.72 , pH = 14-3.92 = 10.284
G) total OH- ions = ( 0.0053 x0.11 + 0.02 x0.092 x 2) = 0.004263
[OH-] = 0.004263/ ( 0.0253) = 0.17
H) pOH = -loh ( 0.17) = 0.77 , pH = 14-0.77 = 13.23
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