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Chapter 0: Q13P (page 1)

How many milliliters of 0.202MNaOHshould be added to 25.0 mlof 0.0233Msalicylic acid (2-hydroxybenzoic acid) to adjust the pHto 3.50?

Short Answer

Expert verified

The required volume of $NaOH is 2.223 mL$

Step by step solution

01

Henderson-Hasselbalch equation

• The Henderson-Hasselbalch equation is a simplified formula that shows the connection between a solution's pH or pOH, pKa or pKb, and the ratio of concentrations of dissociated chemical species.

02

Determine the moles of salicylic acid

Calculate the moles of salicylic acid using the formula below: $η (salicylic acid) = c \times V$

$η (salicylic acid) = (0.023 M) \times (25 mL)$

$η (salicylic acid) = 0.5285 mmo 1$

Consider the reaction:

localid="1663413505585" $H_{2} A + {OH}^{-} ⇌ {HA}^{-} + H_{2} O [H_{2} A] = 0.5285 - x [{HA}^{-}] = x$

03

Determine the required volume of NaOH

The Henderson-Hasselbach equation should be used next. $pH = pKa + \log ([{HA}^{-}]) / [H_{2} A]) 3.50 = 2.972 + \log (\frac{x}{0.5285 - x}) x = 0.4993 mmol$

Calculate the required volume of as follows $V (NaOH) = n \times c V (NaOH) = (0.4493 mmol) \times (0.202 M) V (NaOH) = 2.223 mL$

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Most popular questions from this chapter

Find the area from $800$ to $1000$ hours.

(a) What is the pH at the equivalence point when 0.030 0 M NaF is titrated with 0.060 0 M HClO4?

(b) Why would an indicator endpoint probably not be useful in this titration?

For a solution of ${Ni}^{2 +}$ and ethylenediamine, the following

equilibrium constants apply atrole="math" localid="1663385095626" $20^{°}$ c:

role="math" localid="1663386154645" ${Ni}^{2} + H_{2} {NCH}_{2} {CH}_{2} {NH}_{2} Ni {(cn)}^{2 +} \log K_{1} = 7.25 Eilhylenediamine (abbreviated en) Ni {(en)}^{2} + cn Ni {(en)}_{2}^{2 +} \log K_{2} = 6.32 Ni {(en)}^{2} + cn Ni {(en)}_{3}^{2 +} \log K_{3} = 4.49$

Calculate the concentration of free role="math" localid="1663386355087" ${Ni}^{2 +}$ in a solution prepared by mixing

0.100 mol of en plus 1.00 mL of role="math" localid="1663386332852" $0.0100 {MNi}^{2 +}$ and diluting to 1.00 L with dilute base (which keeps all the enin its unprotonated form). Assume that nearly allnickel is in the form $Ni {(cn)}_{3}^{2 +} so [Ni {(cn)}_{3}^{2 +}] = 1.00 x 10^{- 5} M$ .Calculate the concentrations of

$Ni {(en)}^{2 +} and Ni {(en)}_{3}^{2 +}$ to verify that they are negligible incomparison with $Ni {(en)}_{3}^{2 +}$ .

If there had been $n = 13$ replications in both data sets, would the difference in standard deviations be significant?

What does the designation 200/400mesh mean on a bottle of chromatography stationary phase? What is the size range of such particles? Which particles are smaller,100/200or 200-400mesh?

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