Problem: Calculate the pH of the resulting solution if 27.0 ml of 0.270 M HCl is added toa) 32.0 ml of 0.270 M NaOH(aq)b) 37.0 ml of 0.320 M NaOH(aq)

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a. 32 mL 0.27 M NaOH

mol H+ = (27 mL)(1x10-3 L1 mL)(0.27 mol HCl1 L)(1 mol H+1 mol HCl)mol H+ = 0.00729 molmol OH- = (32 mL)(1x10-3 L1 mL)(0.27 mol NaOH1 L)(1 mol OH-1 mol NaOH)mol OH- = 0.00864 mol


mol OH- excess = 0.00864 mol OH- - 0.00729 H+ = 0.00135 mol OH-

M OH- = 0.00135 mol OH-(32+27)mL(1x10-3 L1 mL) =0.02288 M

pOH = -log(0.02288) = 1.64

pH = 14 - 1.64 = 12.35


b.  37.0 ml of 0.320 M NaOH(aq)

mol OH- = (37 mL)(1x10-3 L1 mL)(0.27 mol NaOH1 L)(1 mol OH-1 mol NaOH)mol OH- = 0.00999 mol

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Problem Details

Calculate the pH of the resulting solution if 27.0 ml of 0.270 M HCl is added to

a) 32.0 ml of 0.270 M NaOH(aq)

b) 37.0 ml of 0.320 M NaOH(aq)

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