Question

Name Section Date QUESTIONS: EXPERIMENT 16 1. What volume of 0.317 M KOH solution will be required to fitrate 25.00 mL of 0.285 M HNO answer 2. If 38.71 mL of 0.108 M NaOH solution is required to titrate a 10.00-ml sample of an unknown H2SO4 solution, what is the molarity of the acid solution? answer 3. What mass of Ca(OH)2 is required to neutralize 50.00 mL of 0.180 M HCI? answer 158

Answer 1

@ At equivalent point equivalent of equivalent of acia Base equivalent = Moles & Valency (v.f.) factor ivalent = Molarity & Volumen - for haioa molarity (ang= 0.285 M Volume (Vi = 95 m2 Vof. = 1 so, equivalent of HNOG = 0285x95x1 = 7•125 meg - Forkor molarity (M) = 0.317M Volume (v) = V mL V.f. = 1 so, equivalent of KOH = 0.317 xVx|| = 0.37v meg

at equivalent point, equivalent of HNO₃ = equivalent of kon 7.12.5 = 0.317 V *V=22:48ML] ②. For Naon M = 0.1081 v= 38.77 ml Nofo=1 equivalent of NaOH= 0.108438.79x1 = 4.18 meq - For H₂SO4 M = M v= 10ml V.F. = 2 equivalent of H₂ SO u = M* 1082 C = (201) meg at equivalent point equivalent of NaOH = equivalent of H₂SO4 4.18 - 20 M

I M = 0.909 in molarity - equivalent = moves & Valency factor 1. = molarity & volumea v For Hel M = 0·180 M V = 50 mL yf.=) equivalent = 0·180x50x1 - 9 meq For Calor2 mous on V.F. - 2 equivalent = n x2 = an so, at equivalent point on = 9 n=4.5 mmols n = 4.5x10 3 moles]

y Moly of Calonls : 4.5*10°3 moll 74 gmimola molar mars = so mass of calohl, = Molest Molor mass = 4.5*10-3x 74 * = 0.333 gm