Chemistry, 06.09.2019 23:30 gayshipsandcoffee
What is the concentration of ch3cooh in 0.40 m n2ch3coo needed to produce a buffer solution with a ph= 4.80? ka (ch3cooh) = 1.8 x 103 - е. 1.80 м с. 2.7 х 10: 3 м a. 0.46 m в. 1.6 х 10* м d. 0.35 m
Answers: 2
![\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\ \\4.80 & = & -\log \left(1.8 \times 10^{-5}\right) + \log\dfrac{0.40}{\text{[HA]}} \\\\4.80 & = & 4.74 + \log\dfrac{0.40}{\text{[HA]}}\\\\0.06 & = & \log\dfrac{0.40}{\text{[HA]}}\\\\1.14 & = & \dfrac{0.40}{\text{[HA]}}\\\\1.14\text{[HA]} & = & 0.40\\\\\text{[HA]} & = & \dfrac{0.40}{1.14}\\\\ & = & \mathbf{0.35}\\\end{array}\\](/tpl/images/0224/8677/ee4e5.png)
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What is the concentration of ch3cooh in 0.40 m n2ch3coo needed to produce a buffer solution with a p...
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