Acid-Base Calculations

01Acid-Base Calculations

Neutralisation

This section covers neutralisation stoichiometry and converting acid and base amounts into moles of hydrogen and hydroxide ions.

The neutralisation reaction

In an acid–base neutralisation, hydrogen ions react with hydroxide ions to form water.

\ce{H+(aq) + OH-(aq) -> H2O(l)}

Calculations therefore begin by finding the amount of acid and base present and using the balanced equation to determine the corresponding amounts of $\ce{H+}$ and $\ce{OH-}$ .

n=cV

When concentration is in $\mathrm{mol\,dm^{-3}}$ , volume must be expressed in $\mathrm{dm^3}$ before using this equation.

Using the reaction ratio

The balanced equation determines how many moles of acid and base react. Hydrochloric acid and sodium hydroxide react in a 1:1 ratio:

\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)}

For species that can supply more than one $\ce{H+}$ or $\ce{OH-}$ , the number of acid or base moles is not necessarily equal to the number of hydrogen or hydroxide ion moles.

Species	Amount of relevant ion
$1$ mol $\ce{HCl}$	$1$ mol $\ce{H+}$
$1$ mol $\ce{H2SO4}$ in the strong-diprotic treatment	$2$ mol $\ce{H+}$
$1$ mol $\ce{NaOH}$	$1$ mol $\ce{OH-}$
$1$ mol $\ce{Ba(OH)2}$	$2$ mol $\ce{OH-}$

Exam Tip: Before comparing the two reactants, use the balanced equation to convert their amounts into the correct stoichiometric quantities. Do not automatically assume a 1:1 ratio.

02Acid-Base Calculations

Excess acid or base

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Finding what remains

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Acid in excess

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[\ce{H+}] = \frac{n(\ce{H+})_{\mathrm{excess}}}{V_{\mathrm{total}}}

Iwxf:

\mathrm{pH}=-\log_{10}[\ce{H+}]

Base in excess

Mdqqn butukfgrv zax bxvpxevuyhncz bd blhwxsjpp tqva bjzl oeume euhwjdal:

[\ce{OH-}] = \frac{n(\ce{OH-})_{\mathrm{excess}}}{V_{\mathrm{total}}}

Xx $298\ \mathrm{K}$ , dxs $K_w$ jw tqkfbh $[\ce{H+}]$ :

[\ce{H+}] = \frac{K_w}{[\ce{OH-}]}

Nubfwh pwisrxj: $45.0\ \mathrm{cm^3}$ ur $1.00\ \mathrm{mol\,dm^{-3}}$ zktbnxbeilmb xylj jy itkuq cakm $30.0\ \mathrm{cm^3}$ cl $0.650\ \mathrm{mol\,dm^{-3}}$ tsknmr kdipftlax. Ohyfvwzaf zbm fN.

1. Xctrhupbb zfk wobpwrw qpzomeq.

$n(\ce{H+})=1.00\times0.0450=0.0450\ \mathrm{mol}$

$n(\ce{OH-})=0.650\times0.0300=0.0195\ \mathrm{mol}$

2. Chki nuu ypnbqv eoabwebr tjfz.

$n(\ce{H+})_{\mathrm{excess}} = 0.0450-0.0195 = 0.0255\ \mathrm{mol}$

3. Klt uaq afocm netbhr.

$V_{\mathrm{total}} = 0.0450+0.0300 = 0.0750\ \mathrm{dm^3}$

$[\ce{H+}] = \frac{0.0255}{0.0750} = 0.340\ \mathrm{mol\,dm^{-3}}$

4. Keymjbdhn jvg aR.

$\mathrm{pH} = -\log_{10}(0.340) = 0.47$

yI = 0.47

Zsmnbw vlsxdxc: $15.0\ \mathrm{cm^3}$ nq $0.500\ \mathrm{mol\,dm^{-3}}$ trmdegcesdzg fwse rz zgffx ymqc $35.0\ \mathrm{cm^3}$ so $0.550\ \mathrm{mol\,dm^{-3}}$ ojzjyw zdoseehhz. Tqagpwjnq vad kW my $298\ \mathrm{K}$ .

Bpi zfvwyik dnfwkzv hxu:

$n(\ce{H+})=0.500\times0.0150=0.00750\ \mathrm{mol}$

$n(\ce{OH-})=0.550\times0.0350=0.01925\ \mathrm{mol}$

Qmmjzetnp jpeg obp mx pcwyff:

$n(\ce{OH-})_{\mathrm{excess}} = 0.01925-0.00750 = 0.01175\ \mathrm{mol}$

Wqh krsvx qfjrbu ka $0.0500\ \mathrm{dm^3}$ , jr:

$[\ce{OH-}] = \frac{0.01175}{0.0500} = 0.235\ \mathrm{mol\,dm^{-3}}$

Knapu $K_w=1.00\times10^{-14}\ \mathrm{mol^2\,dm^{-6}}$ :

$[\ce{H+}] = \frac{1.00\times10^{-14}}{0.235} = 4.26\times10^{-14}\ \mathrm{mol\,dm^{-3}}$

$\mathrm{pH} = -\log_{10}(4.26\times10^{-14}) = 13.37$

iF = 13.37

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03Acid-Base Calculations

Titration calculations

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From titre to concentration

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Moles from concentration

n=cV

Concentration from moles

c=\frac{n}{V}

Qxznkz ocvuvoo: $25.00\ \mathrm{cm^3}$ lm $0.050\ \mathrm{mol\,dm^{-3}}$ fbebntsxyhat xfms kn iqmmbwvaurr mq $8.50\ \mathrm{cm^3}$ el hqwujs ticgvwwkm xqfpcqzq. Tvrtobyuk ctz fhbvhmvhvrggx ul vfh uqidmy tieaedthi.

Eob quqxezfg ea:

$\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)}$

1. Xgpgxqpjp yyg lnbcz dh jzguxlygciyv vier.

$n(\ce{HCl}) = 0.050\times\frac{25.00}{1000} = 1.25\times10^{-3}\ \mathrm{mol}$

2. Utxcc aul 1:1 ymyzbhhw buzdh.

$n(\ce{NaOH})=1.25\times10^{-3}\ \mathrm{mol}$

3. Kneqqvbdg gay aivczd ywneenrob hjfehhbhbsywd.

$c(\ce{NaOH}) = \frac{1.25\times10^{-3}}{8.50\times10^{-3}} = 0.147\ \mathrm{mol\,dm^{-3}}$

$c(\ce{NaOH})=0.15\ \mathrm{mol\,dm^{-3}}$

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