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CH 125 L Lab 14.

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Presentation on theme: "CH 125 L Lab 14."— Presentation transcript:

1 CH 125 L Lab 14

2 Lab 13 – Titration Tutorial
A technique for determining the concentration of a solution. An Erlenmeyer flask is filled with an acid of unknown concentration but a fixed volume A burette is filled with a base of known concentration The base is added to the acid drop by drop till it neutralizes the acid

3 Lab 13 – Titration Tutorial
At the point of neutralization (called END POINT OR EQUIVALENCE POINT OR NEUTRALIZATION POINT): Moles of Acid = Moles of Base Moles = Molarity (Concentration) x Volume So Concentration(acid) x Volume(acid) = Concentration (base) x Volume (base) C(acid) x V(acid) = C(base) x V(base) Have to find C(acid): C(acid) = C(base) x V(base) V(acid)

4 Lab 13- Titration Tutorial
C(base) = Molarity of Base (moles/L) or M(base) C(acid) = Molarity of Acid (moles/L) or M(acid) M(acid) X V(acid) = M(base) x V(base) Have to find M(acid): Where M = Molarity

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6 For an acid base titration the equivalence point is the point where neutralization occurs. The point of neutralization is determined by the color change of the indicator. Indicators must have the following properties : a) the color change must be sharp. i.e. the indicator must change color with the addition of one drop of solution from the burette. b) change color as close to the equivalence point as possible c) Give a distinct color change

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8 Neutralization of Hydrochloric Acid and Sodium Hydroxide
At Neutralization Point the Reaction of NaOH and HCl is complete. They have neutralized each other when: Moles NaOH = Moles HCl The Reaction of Neutralization (or Acid- Base reaction) is: NaOH (aq) + HCl(aq)  NaCl(aq) + H2O(l) high pH low pH pH = 7 (neutral)

9 My data after Conducting Titration
Molarity of NaOH = 0.10M Volume of NaOH in burette at start (mL) = Volume of NaOH in burette at end(mL) = Volume of NaOH delivered into flask (mL) = Volume of HCl in the flask (mL) = 10 C (acid) = C(base) x V(base) = (0.10 M x V(base) = V(acid) mL


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