Problem: Calculate the pH of the resulting solution if 30.0 mL of 0.300 M HCl(aq) is added to(a) 40.0 mL of 0.300 M NaOH(aq).PH=(b) 20.0 mL of 0.400 M NaOH(aq).PH=

FREE Expert Solution

We are being asked to determine the of the resulting solution if 30.0 mL of 0.300 M HCl(aq) is added to


(a) 40.0 mL of 0.300 M NaOH(aq).

Calculate moles: 

mol HCl = 0.300 mol1  L ×30 mL ×10-3 L 1  mL  = 0.009 mol HClmol NaOH = 0.300 mol1  L ×40 mL ×10-3 L 1  mL  = 0.012 mol NaOH


mol NaOH left = 0.012 mol -0.009 mol = 0.003 mol NaOH


NaOH is a strong base so [OH-] = [NaOH]

[OH-]=0.003 mol30 mL+40 mL×1 mL10-3 L[OH-]=0.003 mol70 mL×1  mL10-3 L[OH-]=0.0429 M


Calculate pOH: 

pOH = -log[OH-]pOH = -log[0.0429]

pOH = 1.37


Calculate pH: 

pH = 14-pOHpH = 14-1.37

pH = 12.63


The pH after the titration is 12.63


(b) 20.0 mL of 0.400 M NaOH(aq).

Calculate moles: 

mol HCl = 0.300 mol1  L ×30 mL ×10-3 L 1  mL  = 0.009 mol HClmol NaOH = 0.400 mol1  L ×20 mL ×10-3 L 1  mL  = 0.008 mol NaOH


mol HCl left = 0.009 mol -0.008 mol = 0.001 mol HCl


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Problem Details

Calculate the pH of the resulting solution if 30.0 mL of 0.300 M HCl(aq) is added to

(a) 40.0 mL of 0.300 M NaOH(aq).

PH=


(b) 20.0 mL of 0.400 M NaOH(aq).

PH=

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