Suppose 50.0 mL of 0.300 MHCI is titrated with 0.500 M NaOH. (Remember this mean

Question

Suppose 50.0 mL of 0.300 MHCI is titrated with 0.500 M NaOH. (Remember this means that the 50.0 mL of 0.300 M HCI would be in the flask initially and the buret would contain the 0.500 M NaOH.) a. Calculate the pH in the flask after 20.0 mL of the 0.500 M NaOH is added from the buret. Show your work. b. Calculate the volume of 0.500 M NaOH needed to reach the equivalence point. c. What would be the pH at the equivalence point of this titration? Explain why you would expect the pH to be this particular value. Suppose 25.0 mL of 0.100 M benzoic acid (a weak acid which can be abbreviated HBz) is titrated by 0.200 M NaOH. (K_a of benzoic acid = 6.3 times 10^5) The balanced equation representing the neutralization would be: HBz (aq) + NaOH (aq) rightarrow NaBz (aq) + H_2 O (I) a. Calculate the pH after addition of 4.0 mL of 0.200 M NaOH. Show your work. b. Calculate the pH at the equivalence point of this titration. Show your work. c. Calculate the pH when 5.00 mL of 0.200

Explanation / Answer

18.a) moles of HCl= 0.3*0.05= 0.015

moles of NaOH= 0.5*0.02= 0.010

moles HCl left after addition of 20mL of 0.5M NaOH

0.015-0.010= 0.005moles of HCl

concentration of HCl= 0.005/0.070=0.07143

pH=1.146

b) N1V1=N2V2

0.3*50=0.5*V2

V2=30mL of NaOH

c) since both are strong electrolyte, so they will dissociates completely.

their neutralisation form water and NaCl which neutral in nature.

hence ph should be 7 at equivalence point.

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.