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Chapter 9: Q3TY (page 196)

Find the pH of 0.10 M methylamine.

Short Answer

Expert verified

The pH value for methylamine is 11.80

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.

0.10M methylamine

$K_{b} = \frac{K_{w}}{K_{a}} = \frac{10^{- 14}}{10^{- 10.632}} = 4.3 \times 10^{- 4}$

$K_{b} = \frac{[{CH}_{3} {NH}_{3} {}^{+}H_{2} O]}{[{CH}_{3} {NH}_{2}]} K_{b} = \frac{x^{2}}{0.10 - x} 4.3 \times 10^{- 4} = \frac{x^{2}}{0.10 - X}$

By using quadratic equation solve The value of x .

$x = 6.78 \times 10^{- 3}$

From the x value the is calculated as

$pH = - \log [{OH}^{-}] = - \log (6.78 \times 10^{- 3}) = 2.17 pH + pOH = 14 pH = 14 - 2.17 = 11.83$

Thus the Ph value for methylamine is 11.8

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Most popular questions from this chapter

(a) Calculate the pH of a solution prepared by mixing $0.00100 mol$ of the base $B (K_{b} = 10^{- 2.00})$ with $0.0200 mol$ of ${BH}^{+} {Br}^{-}$ and diluting to $1.00 L$ .First calculate the pH by assuming $[B] = 0.0100$ and $[{BH}^{+}] = 0.0200 M$ . Compare this answer with the pH calculated without making such an assumption.

(b) After working part (a) by hand, use Excel Goal Seck to find the same answers.

Write the chemical reaction whose equilibrium constant is

(a) $K_{a}$ for benzoic acid, $C_{6} H_{5} {CO}_{2} H$

(b) $K_{b}$ for benzoate ion, $C_{6} H_{5} {CO}_{2}^{2 -}$

(c) $K_{b}$ for aniline, $C_{6} H_{5} {NH}_{2}$

(d) $K_{a}$ for anilinium ion, $C_{6} H_{5} {NH}_{3}^{+}$

What is the pH of a solution prepared by dissolving 1.23 of

2-nitrophenol (FM 139.11) in 0.250 L?

A $0.0450 M$ solution of benzoic acid has a pH of $2.78$ . Calculate ${pK}_{a}$ for this acid

${Cr}^{3 +}$ is acidic by virtue of the hydrolysis reaction ${Cr}^{3 +} H_{2} O \overset{K_{aI}}{⇌} Cr {(OH)}^{2 +} + H^{+}$ . [Further reactions produce $Cr {(OH)}_{2,}^{+} Cr {(OH)}_{3}$ and.] Find the value role="math" localid="1654864894349" $K_{aI}$ of in Figure 6-9. Considering only the $K_{aI}$ reaction, find pH value ofrole="math" localid="1654865110221" $0.010 MCr {({CIO}_{4})}_{3}$ What fraction of chromium is in the formrole="math" localid="1654865163335" $Cr {(OH)}^{2 +}$ ?

See all solutions

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