Solution and answer to neutralization problem

When 125.0 mL of 0.25 M HCl is mixed with 75.0 mL of 0.34 M NaOH, what is the nature of the mixture? What is the balanced equation for the reaction between the two solutions? Is the solution neutral, acidic or basic? How many mols of HCl were added? How many moles of NaOH were added? Which is in excess?

This problem requires the balanced equation. The moles of solute in each solution need to be calculated. These mole amounts are compared to the required 1:1 ratio dictated by the balanced equation. hydrochloric acid sodium hydroxide ----> sodium chloride water HCl (aq) + NaOH (aq) - ----> NaCl(aq) + H2O(l) 1 mole acid 1 mole base 1 mole salt 1 mole water The nature of the solution can be decided after the mols of acid and base are calculated. The reaction shows a required 1 mole acid for each 1 mole of base. The mols of acid can be calculated using the definition for molarity moles solute = molarity x liters solution = M V moles acid = Ma x Va The moles of base can be determined using the same method. moles base = Mb x Vb the terms are defined as follows Acid Base Ma is the molarity of the acid Mb is the molarity of the base Va is the volume of the acid Vb is the volume of the base organize data Ma = 0.25 M HCl Mb = 0.34 M NaOH Va = 125.0 mL = 0.125 L Vb = 75.0 mL = 0.075 L substitute values moles of acid = ( 0.25 M HCl )(0.125 L) moles of base = 0.34 M NaOH)(0.075 L) moles of acid = 0.03125 M HCl round to two sf 0.031 moles HCl moles of base = 0.02550 mole NaOH round to two sf 0.026 moles NaOH When the mols of acid and base are compared we see that there are more moles of acid than base. The solution will be acidic.

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