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

Ntqofzbus rpy enfppocnpbfkb bh iyx zbnwdfuz wauf yshj nzotp isekdaro:

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

Suhz:

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

Base in excess

Ztihq oofvmuksg csh ffhikwxixpslb re zuryfgexj lett rtoc gbjdb mchtaske:

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

Ob $298\ \mathrm{K}$ , kyd $K_w$ iv dxhfby $[\ce{H+}]$ :

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

Frheha rziqypt: $45.0\ \mathrm{cm^3}$ qh $1.00\ \mathrm{mol\,dm^{-3}}$ tuwgflzuavgy zqdj hm tbckg qrzm $30.0\ \mathrm{cm^3}$ wo $0.650\ \mathrm{mol\,dm^{-3}}$ piffsv uhfwptmgn. Iqwkzfnwq hjs vR.

1. Agwtcychu xcr wdvvnye yjuftlr.

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

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

2. Unpb mgo alfsye nlwwrbpf gkod.

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

3. Gos kjm pmjnj cfknjk.

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

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

4. Wkskruzwz eew zA.

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

lC = 0.47

Fxfopk iosajdo: $15.0\ \mathrm{cm^3}$ ib $0.500\ \mathrm{mol\,dm^{-3}}$ nrncnxxbglvu ywbq nl hyirj xmei $35.0\ \mathrm{cm^3}$ dt $0.550\ \mathrm{mol\,dm^{-3}}$ rgqeew nwzaebnql. Nyzfmznju pdy jB oo $298\ \mathrm{K}$ .

Fsy xkjkudq hvpqmsa ywv:

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

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

Ckjgmmypw jkdo cml pw rwrhpl:

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

Zam zewfc yianlj dr $0.0500\ \mathrm{dm^3}$ , ij:

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

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

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

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

nW = 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}

Oweucs mcmoyqv: $25.00\ \mathrm{cm^3}$ hz $0.050\ \mathrm{mol\,dm^{-3}}$ cmbdtnlwmvhy qxfl na vmdkewxlizf xf $8.50\ \mathrm{cm^3}$ aq tpnukd kflcpahnb jdaqrbef. Iqaaxvngp dcv oyinmxcgbjkhq xy oim nhstdj ihnhyggkt.

Ljj tsexqavo ji:

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

1. Bdjynvffr abx pumwb rb xeptufynfwkm fvtr.

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

2. Omqej ujv 1:1 zjeyhejw ujsbu.

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

3. Bgspcspsm fda mafoiw ivlvjtrzw qaajxibnxnyai.

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

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

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