In a titration, 20.0 milliliters of 0.15 m hci(aq) is exactly neutralized by 18.0 milliliters of koh(aq).
complete the equation in your answer booklet for the neutralization reaction by writing the formula of each product.
koh(aq) + hcl(aq)==> +
compare the number of moles of h+(aq) ions to the number of moles of oh-(aq) ions in the titration mixture when the hcl(aq) is exactly neutralized by the koh(aq).
determine the concentration of the koh(aq).

The reaction yields H2O and KCl (remember that neutralization reactions produce water and a salt). When HCl is exactly neutralized by KOH, you are left with no excess H+ or OH- from the strong acid or strong base. Also, since both the acid and base are strong, their conjugate base and conjugate acid, respectively, have negligible basicity. Therefore, the final solution is of neutral pH and there are equal amounts of H+(aq) and OH-(aq). Since the acid is monoprotic and the base releases only 1 OH- per molecule dissociated, you need not worry about dissociation constants. In this case, you find the molarity using M1V1 = M2V2. You are finding M2, so M2 = M1V1/V2 = (0.15)(20.0)/(18.0) = 0.17M (this is considering significant figures).

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