Question

Calculate the pH at the halfway point and at the equivalence point for each of the following titrations. 100.0 mL of 0.60 M titrated by 0.10 M pH at the halfway point = pH at the equivalence point = 100.0 mL of 0.70 M titrated by 0.20 M pH at the halfway point = pH at the equivalence point = 100.0 mL of 0.70 M titrated by 0.15 M pH at the halfway point = pH at the equivalence point ='

Answer 1

Answer:

Explanation:

the question is incomplete since the reagents are not given. i will give a similar question and show how to answer it.

Calculate the pH at the halfway point and at the equivalence point for each of the following titrations.

100.0 mL of 0.60 M $C_{6}H_{5}COOH$ ($K_{a}=6.4*10^_{-5}$) titrated by 0.10 M $NaOH$

SOLUTION

pH scale is used to show how acidic or basic a solution is. Also note that for half way titration, the volume of the base used is half of its equivalence point.

to find the volume of the base we use the formula  $M_{1}V_{1}=M_{2}V_{2}$.

M1 is the concentration of the acid  =0.60 M

V1 is the volume of the acid = 100.0 mL =

M2 is thye concentration of the base NaOH =0.10 M

V2 the volume of the base=?

thus we have

0.60 M × 100.0 mL = 0.10 M × V2

V2 = 600 mL(This is the volume of the base at equivalence point)

thus the volume of the base at half way point is 300 mL

the chemical equation of the reaction is given as

$C_{6}H_{5}COOH + NaOH$ → $C_{6}H_{5}COONa + H_{2}O$

0.60 mole        0.005         0.005

using the formula

pH = p$K_{a}$ + log $\frac{salt}{acid}$

     = 4.20 + log $\frac{0.005}{0.005}$  = 4.20

(please note that p$K_{a}$ of benzoic is 4.20   )