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Lina20 [59]
3 years ago
9

A buffer is made that is .10M propanoic acid (pKa = 4.9) and .05M sodium propanoate. Calculate the pH.

Chemistry
1 answer:
inna [77]3 years ago
7 0
PKa= 4.9 therefore ka= 10^-4.9=  1.259x10^-5

ka= \frac{[H^+][CH3CH2COO^-]}{[CH3CH2COOH]}

[CH3CH2COO^-]
= 0.05
[CH3CH2COOH]= 0.10

Therefore 1.259x10^-5 = \frac{[H^+][0.05]}{[0.1]}
Rearrange the equation to make the concentration of hydrogen the subject.
Therefore [H^+] =  \frac{(1.259*10^-5)(0.1)}{0.05}

Therefore [H^+]= 2.513*10^-5

pH= -log [H^+] = -log(2.513*10^-5)= 4.59.
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Explanation:

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ΔT = T_f - T_i = the change in temperature

1. Equation

There are two heat flows in this problem,

heat released by reactor + heat absorbed by water = 0

               q₁                  +                        q₂                     = 0

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q_{1} = -\text{23 746 kJ} \times \dfrac{\text{1000 J}}{\text{1 kJ}} = -\text{23 746 000 J}

(b) ΔT  

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(c) m₂

\begin{array}{rcl}q_{1} + q_{2} & = & 0\\\text{-23 746 000 J} + m_{2} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 85 \, ^{\circ}\text{C} & = & 0\\\text{-23 746 000 J} + 356m_{2} \text{J$\cdot$g}^{-1} & = & 0\\356m_{2} \text{g}^{-1} & = & 23746000\\m_2&=& \dfrac{23746000}{\text{356 g}^{-1}}\\\\ & = & \textbf{67000 g}\\\end{array}\\

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