55.0 mL of 0.500 M lead(II) nitrate (FW = 331.2) is reacted with excess potassium sulfide aqueous solution. Calculate the quantity of lead(II) sulfide (FW = 239.3) precipitated in this reaction. Pb(NO3)2(aq) + K2S(aq)→ PbS(s) + 2 KNO3(aq) O 0.658 g O 6.58 g O 7.61 8 O 3.29 g

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Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.