Answer:
pH = 4.543
Explanation:
- CH3CH2COOH + H2O ↔ CH3CH2COO- + H3O+
- pKa = - Log Ka
∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
∴ pKa = 4.87
⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
added 300 mL 0f 0.02 M NaOH:
⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)
⇒ <em>C</em> CH3CH2COOH = 0.048 M
⇒ <em>C</em> NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M
mass balance:
⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)
charge balance:
⇒ [H3O+] + [Na+] = [CH3CH2COO-]
∴ [Na+] = 0.02 M
⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)
(2) in (1):
⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]
replacing in Ka:
⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])
⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]
⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]
⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0
⇒ [H3O+ ] = 2.867 E-5 M
∴ pH = - Log [H3O+]
⇒ pH = 4.543
Answer:
\large \boxed{\textbf{609 kJ}}
Explanation:
The formula for the heat absorbed is
q = mCΔT
Data:
m = 2.07 kg
T₁ = 23 °C
T₂ = 191 °C
C = 1.75 J·°C⁻¹g⁻¹
Calculations:
1. Convert kilograms to grams
2.07 kg = 2070 g
2. Calculate ΔT
ΔT = T₂ - T₁ = 191 - 23 = 168 °C
3. Calculate q
Answer:
yes your answer is correct for this question.
Answer:
Percent error = 25%
Explanation:
Given data:
Measured density of water = 1.25 g/mL
Accepted density value of water = 1 g/mL
Percent error = ?
Solution:
Formula:
Percent error = (measured value - accepted value / accepted value) × 100
Now we will put the values in formula:
Percent error = (1.25 g/mL - 1 g/mL /1 g/mL )× 100
Percent error = (0.25 g/mL /1 g/mL )× 100
Percent error = 0.25 × 100
Percent error = 25%
