Question

Question 3: Calculate the pH of a solution made by mixing 44.0 mL of 0.044 M HCl and 35.0 mL of 0.055 M RbOH.

1. 5.00

2. 10.11

3. 1.39

4. 8.98

5. 3.86

Question 4: The order of anion stability is correct for which series? (Hint: think about the trend for the conjugate acids.)

i) ^-OCl > ^-OClO > ^-OClO₂ > ^-OClO₃
ii) [H₂SbO₄]^- > [H₂AsO₄]^- > [H₂PO₄]^-
iii) Cl₃CCOO^- > Cl₂CHCOO^- > ClCH₂COO^- > CH₃COO^-
iv) CH₃CH₂COO- > CH₃COO- > Cl₃CCOO- > F₃CCOO-

Answer 1

3)

Given:

M(HCl) = 0.044 M

V(HCl) = 44 mL

M(RbOH) = 0.055 M

V(RbOH) = 35 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.044 M * 44 mL = 1.936 mmol

mol(RbOH) = M(RbOH) * V(RbOH)

mol(RbOH) = 0.055 M * 35 mL = 1.925 mmol

We have:

mol(HCl) = 1.936 mmol

mol(RbOH) = 1.925 mmol

1.925 mmol of both will react

remaining mol of HCl = 1.1*10^-2 mmol

Total volume = 79.0 mL

[H+]= mol of acid remaining / volume

[H+] = 1.1*10^-2 mmol/79.0 mL

= 1.392*10^-4 M

use:

pH = -log [H+]

= -log (1.392*10^-4)

= 3.8562

Answer: 3.86

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