Titrating strong acids and strong bases means that the titrand is strong acid with known concentration and volume, moreover the analyte is strong bases. The reaction during titration is H3O+ + OH- <--> 2H2O. Firstly, we can now start to calculate strong bases(NaOH) volume required for titration. Titrand used for this example is HCl 0.1M and volume 50 mL. From the reaction, we can now that moles of HCl is equal to moles of NaOH, so it could be calculated easily. So, the formula is M1.V1=M2.V2. Which M1 is concentration of HCl and V1 is volume of HCl, also M2 is concentration of NaOH and V2 is volume of NaOH. So, The volume of NaOH required to reach the equivalent point is :
V2 = (M1.V1)/M2 = (0.1 M x 50 mL)/0.2 M = 25 mL
From beginning of titrant added, H+ from HCl is excess in the solution, so the pH level is function of H+. We can calculate it using formula of pH = -log[H3O+] = -log[HCl] = -log[0.1] = 1.
After adding the bases, solution has more NaOH, for example we add 10 mL of NaOH. So, the pH would be :
HCl = [(M1.V1)-(M2.V2)]/V1+V2 = [(0.1 x 50)-(0.2 x 10)]/(50+10) = 0.05 M
pH = -log[HCl] = -log[0.05] = 1.3
At equivalent point, HCl and NaOH has the same moles so it has value of pH same equal to Kw or 10^-7.
So, the value of pH is 7.
After the equivalent point, solution has excess of NaOH than HCl so the equation similar with above formula, just little modification, which in the case of added 30 mL of NaOH :
NaOH [OH-] = [(M2.V2)-(M2.V2)]/V1+V2 = [(0.2 x 30)-(0.1 x 50)]/(50+10) = 0.0125 M
pH is power of H+, so we must calculate of H+ in the solution using formula :
Kw = H+.OH-, H+ = Kw/OH- = 10^-14/0.0125 = 8 x 10^-3
pH = -log[H3O+] = -log[8x10^-3] = 12.10
if we calculate each 5 mL adding of NaOH, the result of curve is like picture above.
V2 = (M1.V1)/M2 = (0.1 M x 50 mL)/0.2 M = 25 mL
From beginning of titrant added, H+ from HCl is excess in the solution, so the pH level is function of H+. We can calculate it using formula of pH = -log[H3O+] = -log[HCl] = -log[0.1] = 1.
After adding the bases, solution has more NaOH, for example we add 10 mL of NaOH. So, the pH would be :
HCl = [(M1.V1)-(M2.V2)]/V1+V2 = [(0.1 x 50)-(0.2 x 10)]/(50+10) = 0.05 M
pH = -log[HCl] = -log[0.05] = 1.3
At equivalent point, HCl and NaOH has the same moles so it has value of pH same equal to Kw or 10^-7.
So, the value of pH is 7.
After the equivalent point, solution has excess of NaOH than HCl so the equation similar with above formula, just little modification, which in the case of added 30 mL of NaOH :
NaOH [OH-] = [(M2.V2)-(M2.V2)]/V1+V2 = [(0.2 x 30)-(0.1 x 50)]/(50+10) = 0.0125 M
pH is power of H+, so we must calculate of H+ in the solution using formula :
Kw = H+.OH-, H+ = Kw/OH- = 10^-14/0.0125 = 8 x 10^-3
pH = -log[H3O+] = -log[8x10^-3] = 12.10
if we calculate each 5 mL adding of NaOH, the result of curve is like picture above.
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