Vibiscms/mod/ibis/view.php?id-5047047 an Sapling Learning Calculate the pH for e

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Vibiscms/mod/ibis/view.php?id-5047047 an Sapling Learning Calculate the pH for each of the following cases in the titration of 35.0 mL of 0.120 M NaOH(aq), with 0.120 M HCI(aq) Number Note: Enter your answers with two decimal places (a) before addition of any HCI Number (b) after addition of 13.5 mL of HCI Number (c) after addition of 22.5 mL of HCI Number (d) after the addition of 35.0 mL of HCl 11D Number (e) after the addition of 43.5 mL of HC0 (1) after the addition of 50.0 mL of HCI O Previous Gwe Up & View Solution e Check Answer Hint Next about uscareers oriväcv policv - terms of use con

Explanation / Answer

both HCl and NaOH are strong acid and base respectively, They undergo complete ionization

volume of NaOH= 35 ml , in terms of liters, the volume= 35/1000 =0.035 L and Normality=0.12M

moles of NaOH= Molarity* volume in L= 0.035*0.12 =0.0042

when no HCl is added, NaOH being strong base, ionizes completely, NaOH ---------->Na++OH-

[OH-]=0.12M, pOH=-log (0.12)= 0.92, pH= 14-pH=14-0.92=13.08

When HCl is added, the following reaction takes place.

NaOH+HCl ----------->NaCl+ H2O, molar ratio of NaOH: HCl= 1:1.

2. moles of HCl added= 0.12*13.5/1000= 0.00162, actual molar ratio of NaOH: HCl=0.0042:0.0016

dividnig by 0.0016, the ratio becomes 0.0042/0.00162 :1= 2.59:1

So excess is NaOH and all the HCl react. so moles of NaOH remaining = 0.0042-0.0016=0.0026

volume of solution after mixing = 35+13.5= 48.5 ml =48.5/1000

concentration of NaOH= 0.0026*1000/48.5 =0.054

[OH-]=0.054, pOH= 1.27, pH= 14-1.27= 12.73

3. moles of HCl added for 22.5ml= 22.5*0.12/1000=0.003, moles of NaOH=0.0042

so ratio of NaOH: HCL=0.0042:0.003= 1.556:1

so excess is NaOH here also and moles of NaOH remaining =0.0042-0.003=0.0012

volume of solution after mixing =35+22.5= 57.5ml

concentration of NaOH= 0.0012*1000/57.5=0.021

[OH-]=0.021, pOH= -log (0.021)=1.68, pH= 14-1.68=12.32

4. when 35 ml of HCl is added, moles of HCl= 35*0.12/1000=0.0042= moles of NaOH

so moles of NaOH: HCL=0.0042:0.0042= 1:1

so both NaOH and HCl completely react and form a neutral salt of pH= 7

5. moles of HCl in 43.5 ml= 0.12*43.5/1000=0.00522, moles of NaOH=0.0042

excess is HCl, moles of HCl remaining after reaction =0.00522-0.0042= 0.00102

volume of solution = 43.5+35= 78.5 ml

concentration of HCl = 0.00102*1000/78.5=0.013

since HCl ionizes completely, [H+]=0.013, pH= 1.88

when 50 ml of HCl is added, moles of HCl= 0.12*50/1000=0.006

here also HCl is excess and moles of HCl remaining = 0.006-0.0042= 0.0018

volume of solution after mixing = 50+35= 85 ml

concentration of HCl after reaction =0.0018*1000/85=0.0211

[H+]=0.0211, pH= -log(0.0211)= 1.67

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