Part E Calculate [OH] for 1.90 mL of 0.140 M NaOH diluted to 1.50 L . Express yo

ID: 1012182 • Letter: P

Question

Part E

Calculate [OH] for 1.90 mL of 0.140 M NaOH diluted to 1.50 L .

Express your answer using three significant figures

Part F

Calculate pH for 1.90 mL of 0.140 M NaOH diluted to 1.50 L .

Express your answer using three decimal places.

Part G

Calculate [OH] for a solution formed by adding 6.00 mL of 0.180 M KOH to 15.0 mL of 8.8×102 M Ca(OH)2.

Express your answer using two significant figures.

Part H

Calculate pH for a solution formed by adding 6.00 mL of 0.180 M KOH to 15.0 mL of 8.8×102 M Ca(OH)2.

Express your answer using two decimal places.

for each and every solution i used ...N1V1 = N2V2

E) N1 = 0.14 M , V1 = 1.9 ml

V2 = 1500 ml

hence N2 = N1V1 /V2

= 0.14 X 1.9 / 1500

= 1.773 X 10-4 M

[OH-] =1.773 X 10-4

hence pOH = 3.751

or pH = 14-pOH = 14-3.751

= 10.249

G) normality of Ca(OH)2 = 2X molarity (because of presence of 2 furnishable OH-)

total gram equivalents of [OH-] = gram equivalents of [OH-] from NaOH + gram equivalents of [OH-] from Ca(OH)2

N1V1 + N2V2

= 6 X 0.18 +15 X 2 X 8.8 X 10-2

= 3.72

or concentation of [OH-] = total gram equivalents / total volume

=3.72 / (6+15)

0.177 M

H) since [OH-] = 0.177 M

we have pOH = -log[OH-]

= 0.75

or pH = 14-pOH = 13.25

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