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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 → 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

A 325-mL sample of solution contains \(25.3 \mathrm{~g}\) of \(\mathrm{CaCl}_{2}\). (a) Calculate the molar concentration of \(\mathrm{Cl}^{-}\) in this solution. (b) How many grams of \(\mathrm{Cl}^{-}\) are in \(0.100 \mathrm{~L}\) of this solution?

Predict and explain which of the following systems are electrically conducting: (a) solid \(\mathrm{NaCl}\), (b) molten \(\mathrm{NaCl}\), (c) an aqueous solution of \(\mathrm{NaCl}\)

Calculate the mass of precipitate formed when \(2.27 \mathrm{~L}\) of \(0.0820 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) are mixed with \(3.06 \mathrm{~L}\) of \(0.0664 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\)

A 0.8870 -g sample of a mixture of \(\mathrm{NaCl}\) and \(\mathrm{KCl}\) is dissolved in water, and the solution is then treated with an excess of \(\mathrm{AgNO}_{3}\) to yield \(1.913 \mathrm{~g}\) of \(\mathrm{AgCl}\). Calculate the percent by mass of each compound in the mixture.

A sample of \(0.6760 \mathrm{~g}\) of an unknown compound containing barium ions \(\left(\mathrm{Ba}^{2+}\right)\) is dissolved in water and treated with an excess of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\). If the mass of the \(\mathrm{BaSO}_{4}\) precipitate formed is \(0.4105 \mathrm{~g}\), what is the percent by mass of Ba in the original unknown compound?
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