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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.

When is a weak acid weak and when is a weak acid strong? Show that the weak acid $99 %$ dissociatedrole="math" localid="1654854665468" $F = (0.0102) K_{a}$ will be 92%dissociated when dissolved in water if the formal concentration is one-tenth $(F = Ka / 10)$ of Show that the fraction of dissociation is 27%when $F = 10 Ka$ . At what formal concentration will the acid be 99%dissociatedCompare your answer with the left -hand curve in Figure 9-2

Which of the following bases would be most suitable for preparing a buffer of $pH 9.00$ ?

(i) ${NH}_{3}$ (ammonia, $K_{b} = 1.76 \times 10^{- 5}$ );

(ii) role="math" localid="1654764490555" $C_{6} H_{5} {NH}_{2}$ (aniline,role="math" localid="1654764515634" $C_{6} H_{5} {NH}_{2}$ );

(iii) $H_{2} {NNH}_{2} (hydrazine, K_{b} = 1.05 \times 10^{- 6}))$ );

(iv) $C_{6} H_{s} {NH}_{2}$ (pyridine, $K_{b} = 1.58 \times 10^{- 9})$ ).

How many mL of 0.500 M NaOH should be added to 10.0 g of tris hydrochloride to give a pH of 7.40 in a final volume of 500 mL?

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

See all solutions

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