A sample of 1.029g of KC8H5O4 was titrated with an NAOH solution of unknown concentration. If 22.83 mL of base was used to reach the endpoint, what was the concentration of NaOH? What mass of KC8H5O4 should be used such that a volume of NaOH between 35-40 mL is needed to reach the endpoint?
2007-11-01 14:24:57 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Atomic weights: K=39 C=12 H=1 O=16 KC8H5O4=204
Let KC8H5O4 be called KP. Let NaOH solution be called S.
1.029gKP/22.83mLS x 1molKP/204gKP x 1molNaOH/1molKP x 1000mLS/1LS = 0.2209 mole NaOH per L, which is molarity.
Suppose 37 mL S is desired.
37mLS x 0.2209molNaOH/1000mLS x 1molKP/1molNaOH x 204gKP/1molKP = 1.67g KP
2007-11-01 14:39:15 · answer #1 · answered by steve_geo1 7 · 0⤊ 0⤋
Mwt potassium hydrogen phthlate = 204.22 (KHP)
[NaOH] = 1000 x wt KHP / (Titre x 204.22)
wt KHP = 35 x 204.22/1000 to 40 x 204.22/1000
U can do the arith.
2007-11-01 21:41:16 · answer #2 · answered by Aurium 6 · 0⤊ 0⤋