![\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.04}{0.08}\right )\\\\& = & 2.865 + \log0.50 \\& = &2.865 - 0.30 \\& = & \mathbf{2.56}\\\end{array}](/tpl/images/0610/7910/ebc2c.png)
![\text{[H$^{+}$]} = 10^{-\text{pH}} \text{ mol/L} = 10^{-2.56}\text{ mol/L} = 2.73 \times 10^{-3}\text{ mol/L}](/tpl/images/0610/7910/adeed.png)
![I = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\I = \dfrac{1}{2}\left [0.04\times (+1)^{2} + 0.04\times(-1)^{2} + 0.00273\times(+1)^{2}\right]\\\\= \dfrac{1}{2} (0.04 + 0.04 + 0.00273) = \dfrac{1}{2} \times 0.08273 = 0.041](/tpl/images/0610/7910/9dfcb.png)
![\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{a_{\text{A}^{-}}}{a_{\text{[HA]}}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.032}{0.08}\right )\\\\& = & 2.865 + \log0.40 \\& = & 2.865 -0.40\\& = & \mathbf{2.46}\\\end{array}\\](/tpl/images/0610/7910/7d363.png)
![I = \dfrac{1}{2}\left [0.10\times (+1)^{2} + 0.05 \times(-1)^{2} + 0.05\times(-1)^{2}\right]\\\\= \dfrac{1}{2} (0.10 + 0.05 + 0.05) = \dfrac{1}{2} \times 0.20 = 0.10](/tpl/images/0610/7910/85fb3.png)
We mix 0.08 moles of chloroacetic acid (ClCH2COOH) and 0.04 moles of
sodium chloroacetate (ClC...
We mix 0.08 moles of chloroacetic acid (ClCH2COOH) and 0.04 moles of
sodium chloroacetate (ClCH2COONa) in 1.0 L of water (pKa = 2,865).
to. Calculate the pH
yes. Calculate the pH using the formal forms (activities). Have on
counts the contribution of the protons (section a) in the calculation of the ionic strength.
C. Find the pH of a mixture prepared by dissolving the following compounds
in a final volume of 1L: 0.08 moles of ClCH2COOH, 0.04 moles of
ClCH2COONa, 0.05 moles of HNO3 and 0.06 moles of NaOH
Answers: 2
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