A 0.050 M solution of the salt NaA has a pH of 9.00. Determine the [OH]and the pH of a 0.010 M solution of the acid HA. Show your work in thequestion parts below.Consider the ionization equation described below.A (aq) + H20(1) = HA(aq) + OH (aq)1Drag the tiles into the numerator or denominator to form the expression for Kb..RESET[0][x][0.050][0.100][1.0 x 10-9)[1.0 x 10-5]1.0 x 10-142.0 x 10-9[0.050 + 1.0x10-](0.050 + 1.0x10-91[0.100 + 1.0x10-1[0.100 + 1.0x10-1[0.050 - 1.0 x10-51[0.050 - 1.0 x10-91[0.100 - 1.0 x10-$] [0.100 - 1.0 ×10-]

Initial pH = 9.0 Calculation of [H+]: pH=-log [H+]9=-log…

Answered: A 0.050 M solution of the salt NaA has…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Step 1

Initial pH = 9.0

Calculation of [H⁺]:

$\begin{array}{rcl} pH & = & - \log [H^{+}] \\ 9 & = & - \log [H^{+}] \\ [H^{+}] & = & 10^{- 9} \\ [{OH}^{-}] & = & \frac{K_{w}}{[H^{+}]} \\ = & \frac{1.0 \times 10^{- 14}}{10^{- 9}} \\ = & 1.0 \times 10^{- 5} M \end{array}$

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