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Chapter 4: Problem 78

Calculate the concentration (in molarity) of a \(\mathrm{NaOH}\) solution if \(25.0 \mathrm{~mL}\) of the solution are needed to neutralize \(17.4 \mathrm{~mL}\) of a \(0.312 \mathrm{M} \mathrm{HCl}\) solution.

Short Answer

Expert verified
The concentration of the NaOH solution is 0.217 M.

Step by step solution

01

Identifying the balanced chemical equation

We first need to write down the balanced chemical equation for the reaction between \(NaOH\) and \(HCl\). In this reaction, NaOH reacts with HCl to produce NaCl and H2O.\n\nNaOH + HCl 鈫� NaCl + H2O\n\nFrom the balanced equation, it is clear that one mole of NaOH reacts with one mole of HCl, indicating that the molar ratio between NaOH and HCl is 1:1.
02

Calculate the number of moles of HCl

The problem provides volume of HCl (in milliliters) and its molarity (M). The number of moles of a solution can be found using the formula:\n\nNumber of moles = Molarity x Volume (in L)\n\nSo, we calculated number of moles of HCl needed for the reaction to be:\n\nNumber of moles of HCl = 0.312 M x 17.4 ml (or 0.0174 L) = 0.0054248 moles.
03

Calculate the molarity of NaOH

Having calculated number of moles of HCl and knowing that the molar ratio between NaOH and HCl is 1:1, we infer that same number of moles of NaOH is required for neutralization.\n\nMolarity is defined as the number of moles of solute per litre of solution. It can be calculated using the formula:\n\nMolarity = Number of moles/volume (in L)\n\nUsing the above formula, we find the concentration (Molarity) of NaOH = 0.0054248 moles/0.025 L = 0.216992 M (or to three significant figures as 0.217 M)

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Most popular questions from this chapter

Arrange the following species in order of increasing oxidation number of the sulfur atom: (a) \(\mathrm{H}_{2} \mathrm{~S}\) (b) \(\mathrm{S}_{8}\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (d) \(\mathrm{S}^{2-},(\mathrm{e}) \mathrm{HS}^{-},(\mathrm{f}) \mathrm{SO}_{2},(\mathrm{~g}) \mathrm{SO}_{3}\)

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Write ionic and net ionic equations for the following reactions: (a) \(\mathrm{Na}_{2} \mathrm{~S}(a q)+\mathrm{ZnCl}_{2}(a q) \longrightarrow\) (b) \(\mathrm{K}_{3} \mathrm{PO}_{4}(a q)+3 \mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow\) (c) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow\)

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