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Chapter 9: Q2TY (page 194)

Find the pH of 0.050 M triethylammonium bromide.

Short Answer

Expert verified

The pH value is 6.01

Step by step solution

01

Step 1: Define the term pH.

The negative logarithm of H+ ion concentration is used to calculate pH.

As a result, the meaning of pH is justified as hydrogen power.

02

Calculate the pH

$Given 0.050 M triethylammonium bromide K_{a} = 10^{- {pK}_{a}} = 10^{- 10.72} = 1.9 \times 10^{- 11} {(C_{2} H_{5})}_{3} {NH}^{+} \overset{K_{a}}{𝆏 {(C_{2} H_{5})}_{3} N + H^{+}}^{0.050 - X X X at equilibrium} \frac{[{(C_{2} H_{5})}_{3} N] [H^{+}]}{[{(C_{2} H_{5})}_{3} {NH}^{+}]} = K_{a} \frac{x^{2}}{0.050 - x} = 1.9 \times 10^{- 11} A quadratic equation is used to solve the value of x . x = 9.75 \times 10^{- 7} From the x value the is calculated as pH = - \log [H_{3} O^{+}] = - \log (9.75 \times 10^{- 7}) = 6.01 Thus the PH value is 6.01$

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