When a small amount of acid is added to a non-buffered solution, there is a larg

Question

When a small amount of acid is added to a non-buffered solution, there is a large change in pH. Calculate the pH when 21.9 mL of 0.0038 M HCl is added to 100.0 mL of pure water. Comment and hint in the general feedback. A buffer solution that is 0.10 M sodium acetate and 0.20 M acetic acid is prepared. Calculate the initial pH of this solution. The K_a for CH_3COOH is 1.8 times 10^-5 M. As usual, report pH to 2 decimal places. A buffered solution resists a change in pH. Calculate the pH when 20.5 mL of 0.039 M HCl is added to 100.0 mL of the above buffer. Calculate how many mL of 0.100 M NaOH are needed to neutralize completely 17.0 mL of 0.0200 M HCl.

Explanation / Answer

6) pure water is naturally 1.00 x 10¯7 M in hydrogen ion.

Calculate total H+ in 100 mL of water:

(1.00 x 10¯7 mol/L) (0.1 L) = 1.0x10-8 mol

Calculate total H+ in HCl:

(0.0038 mol/L) (0.0219 L) = 0.00008322 mol

Add the results together:

0.00008322 mol + 1.0x10-8 mol = 0.00008323 mol

Calculate new molarity of hydrogen ion:

0.00008323 mol / 0.1219 L = 0.00068278 M

Calculate new pH:

pH = - log 0.00068278 = 3.166

7) To get the pH of the buffer solution, we should use the Henderson-Hasselbalch equation which states that

CH3COOH(aq) + H2O(l) --> H3O+(aq) + CH3COO-(aq)

[H3O+] = Ka[CH3COOH] / [CH3COO-]

let if we take 100 ml of both solution

(0.20 M CH3COOH)(0.1 L) = 0.02 mol CH3COOH

(0.1 M NaCH3COO)(0.1 L) = 0.01 mol NaCH3COO

[H3O+] = (1.8 x 10-5) (0.0200/0.01) = 3.6 x 10-5

pH =4.44

8) First, write the equation for the ionization of acetic acid in water and the related Ka expression rearranged to solve for the hydronium ion concentration.

CH3COOH(aq) + H2O(l) --> H3O+(aq) + CH3COO-(aq)

[H3O+] = Ka[CH3COOH] / [CH3COO-]

CH3COOH .......H3O+ ............CH3COO-

initial ...0.02 moles 0.0007995 moles .... 0.01 moles

change + 0.0007995 moles -0.0007995 ..... -0.0007995

eq. 0.0207995 .... 0.000 ..... 0.0092 moles

[H3O+] = Ka [CH3COOH] /[CH3COO-]

[H3O+] = 1.8x10-5 [ 0.0207995] / [0.0092]

[H3O+] = 4.069 x10-5 M

pH = -log[H3O+]

pH = 4.39 ~ 4.4

9) balance chemical reaction

NaOH + HCl ----> NaCl + H2O

1 mole of NaOH react with 1 mole of HCl for neutralize equal no. of mole required

so,

moles of HCl = 0.0200 M x 0.017 L = 0.00034 moles

volume of NaOH = no. of moles / molarity

volume of NaOH = 0.00034 moles / 0.1 M = 0.0034 L = 3.4 mL

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