Question

In: Chemistry

Part A Calculate [OH−] for 1.20 mL of 0.220 M NaOH diluted to 1.50 L Part...

Part A Calculate [OH−] for 1.20 mL of 0.220 M NaOH diluted to 1.50 L

Part B.Calculate [OH−] for a solution formed by adding 4.60 mL of 0.120 M KOH to 14.0 mL of 8.1×10−2 M Ca(OH)2

Thank you!!

Expert Solution

before dilution After dilution

M1 = 0.22M M2 =

V1 = 1.2ml V2 = 1.5L = 1500ml

M1V1 = M2V2

M2 = M1V1/V2

= 0.22*1.2/1500 = 0.000176M

POH = -log[OH-]

= -log0.000176

= 3.7544

PH = 14-POH

= 14-3.7544

= 10.2456

part-B

no of moles of KOH = molarity * volume in L

= 0.12*0.0046 = 0.000552 moles

no of moles of Ca(OH)2 = molarity * volume in L

= 0.081*0.014 = 0.001134 moles

Ca(OH)2 -------------------> Ca^+2(aq) + 2OH^-

0.001134 moles 2*0.001134 moles

Total no of moles of OH^- = 0.000552 + 2*0.001134 = 0.00282 moles

Total volume in L = 0.0046 + 0.014 = 0.0186 L

molarity OH^- = no of moles/volume in L

= 0.00282/0.0186 = 0.1516M

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