Calculate the pH of the resulting solution if 35.0...

Question

Calculate the pH of the resulting solution if 35.0 mL of 0.350 M HCl(aq) is added toa) 25.0 mL of 0.450 M NaOH(aq)ph=____b) 45.0 mL of 0.350 M NaOH(aq).ph= ______

Jules Bruno · Accepted Answer

Let’s first write the balanced reaction between HCl and NaOH:▪ HCl → strong binary acid▪ NaOH → (OH- with Group 1A ion) → strong base▪ the reaction between a strong base and strong acid → no need to create an ICE chartBalanced reaction:              HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)a) 35.0 mL of 0.350 M HCl(aq) is added to 25.0 mL of 0.450 M NaOH(aq)Step 1: Calculate the initial amount (in moles) of the HCl and NaOH [readmore]Recall:molarity (M)=molLmoles HCl=0.350 mol HClL 35.0 mL x 10-3 L1 mLmoles HCl = 0.01225 mol HClmoles NaOH=0.450 mol KOHL25.0 mL x 10-3 L1 mLmoles NaOH = 0.01125 mol NaOHThe amount of HCl is greater than NaOH, therefore there will be HCl left after the reactionStep 2: Calculate the amount of HCl reactedBalanced reaction:                 HCl + NaOH → H2O(l) + NaCl(aq)• 1 mole of HCl reacts with 1 mole of NaOH:• mol of HCl reacted = mol NaOH reactedmoles HCl reacted = 0.01125 mol HClStep 3: Calculate the amount of HCl left after the reactionmoles HCl left = moles HCl (initial) – moles HCl (reacted)moles HCl left = 0.01225 mol HCl - 0.01125 mol HClmoles HCl left = 0.001 mol HClStep 4: Calculate the concentration of HCl left in the solutionmolarity (M)=molLfinal volume of solution=volume HCl +volume KOHfinal volume of solution=35.0 mL x 10-3 L1 mL+25.0 mL x 10-3 L1 mLfinal volume of solution=0.035 L+0.025 LFinal Volume of solution = 0.06 Lmolarity=0.001 mol HCl0.06 Lmolarity=0.0167 mol HClLmolarity = 0.0167 M HClStep 5: Find pH of the solution• When getting the pH of a strong acid solution:pH=-log[strong acid]pH=-log[HCl] pH=-log[0.0167 M]pH = 1.78b) 35.0 mL of 0.350 M HCl(aq) is added to 45.0 mL of 0.350 M NaOH(aq).Step 1: Calculate the initial amount (in moles) of the HCl and NaOH Recall:molarity (M)=molLmoles HCl=0.350 mol HClL 35.0 mL x 10-3 L1 mLmoles HCl = 0.01225 mol HClmoles NaOH=0.350 mol KOHL45.0 mL x 10-3 L1 mLmoles NaOH = 0.01575 mol NaOHThe amount of NaOH is greater than HCl, therefore there will be NaOH left after the reactionStep 2: Calculate the amount of NaOH reactedBalanced reaction:                 HCl + NaOH → H2O(l) + NaCl(aq)• 1 mole of HCl reacts with 1 mole of NaOH:• mol of NaOH reacted = mol HCl reactedmoles NaOH reacted = 0.01225 mol NaOHStep 3: Calculate the amount of NaOH left after the reactionmoles NaOH left = moles NaOH (initial) – molesNaOH (reacted)moles NaOH left = 0.01575 mol NaOH - 0.01225 mol NaOHmoles NaOH left = 0.0035 mol NaOHStep 4: Calculate the concentration of NaOH left in the solutionmolarity (M)=molLfinal volume of solution=volume HCl +volume KOHfinal volume of solution=35.0 mL x 10-3 L1 mL+45.0 mL x 10-3 L1 mLfinal volume of solution=0.035 L+0.045 LFinal Volume of solution = 0.08 Lmolarity=0.0035 mol NaOH0.08 Lmolarity=0.04375 mol NaOHLmolarity = 0.04375 M HClStep 5: Find pOH of the solution• When getting the pOH of a strong base solution:pOH=-log[strong base]pOH=-log[NaOH] pOH=-log[0.04375 M]pOH = 1.36Step 6: Find pH of the solutionpH + pOH =14pH =14 - pOHpH =14 - 1.36pH = 12.64

Problem: Calculate the pH of the resulting solution if 35.0 mL of 0.350 M HCl(aq) is added toa) 25.0 mL of 0.450 M NaOH(aq)ph=b) 45.0 mL of 0.350 M NaOH(aq).ph= __

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Problem: Calculate the pH of the resulting solution if 35.0 mL of 0.350 M HCl(aq) is added toa) 25.0 mL of 0.450 M NaOH(aq)ph=____b) 45.0 mL of 0.350 M NaOH(aq).ph= ______

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Problem: Calculate the pH of the resulting solution if 35.0 mL of 0.350 M HCl(aq) is added toa) 25.0 mL of 0.450 M NaOH(aq)ph=b) 45.0 mL of 0.350 M NaOH(aq).ph= __