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Chapter 6: Q4TY (page 132)

Find $[{OH}^{-}]$ iflocalid="1663407699090" $[H^{+}] = 1.0 脳 10^{- 4} M [H^{+}] = 1.0 脳 10^{- 4} M$

Short Answer

Expert verified

Thus, the value for $[{OH}^{-}]$ was calculated as $1.0 脳 10^{- 10} M$ .

Step by step solution

01

Define The formula for finding .

The value of $K_{b}$ can be calculated as:

$K_{b} = \frac{[{BH}^{+}] [OH]}{[B]}$

Where $K_{b}$ is called the base dissociation constant.

02

Step 2: Calculate the value for [OH-]

Given:

$[H^{+}] = 1.0 脳 10^{- 4} M$

Concentration of $[O H^{-}]$

$K_{w} = K_{a} 脳 K_{b}$

Where,

$(K_{b})$ is the base dissociation constant.

$(K_{w})$ is the ionic product of water $(1.00 脳 10^{- 14})$

$(K_{w})$ is the acid dissociation constant.

$K_{w} = [H^{+}] 脳 [O H^{-}]$

$1.0 脳 10^{- 14} = 1.0 脳 10^{- 4} M 脳 [{OH}^{-}]$

$[O H^{-}] = 1.0 脳 10^{- 10} M$

Thus the value for was calculated as $1.0 脳 10^{- 10} M$

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Most popular questions from this chapter

Write the $K_{b}$ reaction for pyridine and for sodium 2-mercaptoethanol.

: Write thereaction for pyridine and for sodium 2-mercaptoethanol.

Although $KOH, RbOH$ and CsOHhave little association between metal and hydroxide in aqueous solution, ${Li}^{+}$ and ${Na}^{+}$ do form complexes with:

role="math" localid="1663390733265" ${Li}^{+} + {OH}^{-} 鈬� LiOH (aq) K_{1} = \frac{[LiOH (aq)]}{[{Li}^{+}] [{OH}^{-}]} = 0.83$

${Na}^{+} + {OH}^{-} 鈬� NaOH (aq) K_{1} = 0.20$

Prepare a table like the one in Exercise 6-B showing initial and final Concentrations of ${Na}^{+} + {OH}^{-} 鈬� NaOH (aq)$ and (aq)insolution. Calculate the fraction of sodium in the form (aq)at equilibrium.

Find $[{La}^{3 +}]$ in the solution when excess solid lanthanum iodate, $La {({IO}_{3})}_{3}$ , is stirred with until the system reaches equilibrium. Assume that ${IO}_{3}^{-}$ from ${La}_{3} {({IO}_{3})}_{3}$ is negligible compared with ${IO}_{3}^{-}$ from ${LiOO}_{3}$ .

$K_{a}$ for chloroacetic acid is $1.36 脳 10^{- 3} .$ Find $K_{b}$ for chloroacetate ion.

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