Calculate [ O H ? ] for 1.70 m L of 0.160 M N a O H diluted to 1.50 L Part E Cal

Question

Calculate [OH ?] for 1.70mL of 0.160M NaOH diluted to 1.50L Part E Calculate [OH ?] for 1.70mL of 0.160M NaOH diluted to 1.50L . Express your answer using three significant figures. [OH ?] =   M   SubmitMy AnswersGive Up Part F Calculate pH for 1.70mL of 0.160M NaOH diluted to 1.50L . Express your answer using three decimal places. pH = SubmitMy AnswersGive Up Part G Calculate [OH ?] for a solution formed by adding 6.00mL of 0.130M KOH to 11.0mL of 6.0 Calculate [OH ?] for 1.70mL of 0.160M NaOH diluted to 1.50L Part E Calculate [OH ?] for 1.70mL of 0.160M NaOH diluted to 1.50L . Express your answer using three significant figures. [OH ?] =   M   SubmitMy AnswersGive Up Calculate [OH ?] for 1.70mL of 0.160M NaOH diluted to 1.50L Part E Calculate [OH ?] for 1.70mL of 0.160M NaOH diluted to 1.50L . Express your answer using three significant figures. [OH ?] =   M   SubmitMy AnswersGive Up [OH ?] =   M   [OH ?] =   M   SubmitMy AnswersGive Up Part F Calculate pH for 1.70mL of 0.160M NaOH diluted to 1.50L . Express your answer using three decimal places. pH = SubmitMy AnswersGive Up Part F Calculate pH for 1.70mL of 0.160M NaOH diluted to 1.50L . Express your answer using three decimal places. pH = SubmitMy AnswersGive Up pH = pH = SubmitMy AnswersGive Up Part G Calculate [OH ?] for a solution formed by adding 6.00mL of 0.130M KOH to 11.0mL of 6.0 Part G Calculate [OH ?] for a solution formed by adding 6.00mL of 0.130M KOH to 11.0mL of 6.0 [OH ?] =   M  

Explanation / Answer

(E) NaOH => Na+ + OH-

Moles of NaOH = 1.70/1000 x 0.160 = 0.000272 mol


[OH-] = [NaOH] = moles/volume of NaOH

= 0.000272/1.50

= 0.000181 M = 1.81 x 10^(-4) M


(F) pOH = -log[OH-] = -log(1.81 x 10^(-4)) = 3.742

pH = 14 - pOH = 14 - 3.74 = 10.258


(G) KOH => K+ + OH-

Ca(OH)2 => Ca2+ + 2 OH-


Moles of KOH = 6.00/1000 x 0.130 = 0.00078 mol

Moles of Ca(OH)2 = 11.0/1000 x 6.0 x 10^(-2) = 0.00066 mol

Total volume = 6.00 + 11.00 = 17.00 mL = 0.017 L


[OH-] = [KOH] + 2 x [Ca(OH)2]

= (moles of KOH + 2 x moles of Ca(OH)2)/total volume

= (0.00078 + 2 x 0.00066)/0.017

= 0.124 M = 0.12 M


(H) pOH = -log[OH-] = -log(0.124) = 0.91

pH = 14 - pOH = 14 - 0.91 = 13.09

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