A 1.00 I, buffer solution is 0.250 M in Hf and 0.250 M in NaF. Calculate the pH

Question

A 1.00 I, buffer solution is 0.250 M in Hf and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.150 moles of solid NaOH. Assume so volume change upon the addition of base. The K_a for HF is 3.5 times 10^4 A) 3.46 B) 4.06 C) 2.85 D) 3.63 E) 4.24 A 100.0 mL sample of 0.18 M HCIO is nitrated with 0.135 M Ba (OH) Determine the pH of the solution after the addition of 66.67 mL of Ba(OH)_2 (this is the equivalence point) A) -2 B) 7 E) 16 Determine the molar solubility of BaF_2 in pure water K for BaF_2 = 2.45 times 10^-7 A) 1.83 times 10^-2 M B) 1.23 times 10^-5 M C) 2.90 times 10^-2 M D) 4.95 times 10^-3 M E) 6.13 times 10^-6 M Which of the following is an expression for a type of equilibrium constant? A) K_sp and K_f B) K_a and K_b C) pK_a and pK_b D) K_H E) all of the above

Explanation / Answer

Moles of NaOH added = 0.150 moles

This NaOH reacts with HF to form NaF and H2O

So, total moles of NaF = 0.150 moles + 0.250 moles = 0.4 moles

Moles of HF left = 0.250-0.150 = 0.10 moles

Now,

pH =pKa+ log (NaF/HF)

= 3.455 + log(0.4/0.1)

= 4.06

Thus, pH of the solution is 4.06

(14)

Moles of HClO = 0.1 L x 0.18 M = 0.018 moles

Moles of Ba(OH)2 = 0.0667 L x 0.135 M ==0.009 moles

The reaction is,

2 HClo + Ba(OH)2 = Ba(Clo)2 + 2 H2O

So, moles of HClO and Ba(OH)2 left will be zero.

HClO is weak acid, pH is less than 7

PLS post remaining questions separately. According to guidelines only 1 should be answered but i answered 2 to help u

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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