A student neutralized 16.4 milliliters of HCl by adding 12.7 milliliters of 0.620 M KOH. What was the molarity of the HCl acid?
1 answer:
V ( HCl ) = 16.4 mL / 1000 => 0.0164 L M( HCl) = ? V( KOH) = 12.7 mL / 1000 => 0.0127 L M(KOH) = 0.620 M Number of moles KOH: n = M x V n = 0.620 x 0.0127 n = 0.007874 moles of KOH number of moles HCl : <span>HCl + KOH = H2O + KCl </span> 1 mole HCl ------ 1 mole KOH <span>? mole HCl--------0.007874 moles KOH </span> moles HCl = 0.007874 * 1 / 1 = 0.007874 moles of HCl M = n / V M = 0.007874 / <span>0.0164 </span>= 0.480 M Answer (2) hope this helps!
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