Part E Calculate [OH] for 1.50 mL of 0.180 M NaOH diluted to 1.50 L . Express yo

Question

Part E

Calculate [OH] for 1.50 mL of 0.180 M NaOH diluted to 1.50 L .

Express your answer using three significant figures.

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Part F

Calculate pH for 1.50 mL of 0.180 M NaOH diluted to 1.50 L .

Express your answer using three decimal places.

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Part G

Calculate [OH] for a solution formed by adding 6.00 mL of 0.130 M KOH to 17.0 mL of 9.8×102 M Ca(OH)2.

Express your answer using two significant figures.

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Part H

Calculate pH for a solution formed by adding 6.00 mL of 0.130 M KOH to 17.0 mL of 9.8×102 M Ca(OH)2.

Express your answer using two decimal places.

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Explanation / Answer

Part E:

Initial volume of NaOH = 1.50 mL

Initial concentration of NaOH = 0.180 M

Final volume = 1.50 L = 1500 mL

We know that:

M1V1 = M2V2

0.180 * 1.50 = M2 * 1500

M2 = 0.00018

Hence [OH-] = 0.00018 M

Part F:

We know that;

pOH = - log [OH-]

= - log 0.00018

= 3.74

pH = 14 - pOH

= 14 - 3.74

pH = 10.26

Part G:

6.00 mL of 0.130 M KOH and,

17.0 mL of 9.8×102 M Ca(OH)2

Milimoles of KOH = 6 * 0.130

= 0.78

So milimoles of OH from KOH = 0.78

Milimoles of Ca(OH)2 = 9.8×102 * 17

= 1.666

Milimoles of OH- from Ca(OH)2 = 2 * 1.666

= 3.332

Total milimoles of OH- = 3.332 + 0.78

= 4.112

Total volume = 6 + 17 = 23 mL

[OH-] = 4.112 / 23

= 0.178 M

Part H:

[OH-] = 0.178 M

pOH = - log (0.178)

= 0.75

pH = 14 - 0.75

= 13.25

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