Question

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We dissolve 3.44 g of an unknown acid, HA, in enough water to produce 25.0 mL of solution. We titrate this solution with a 0.250 M solution of NaOH. It takes 18.3 mL of the NaOH solution to reach the equivalence point. The pH at the equivalence point is 12.00. (a) (3 points) What is the molar mass of HA? (b) (3 points) What is the pK, value of HA(aq)? (c) (3 points) What was the original pH of the solution of HA(aq)?

Answer 1

Giney : 3.44g of HA (via 25 m ha mL) + 0.250 M Neoh Narom18.3 m) at equing lencept 12.00 HA man nofor pH at Equina lence =) (a) of - At equina lence ? mores of pt. moles of Ngon HA 3.449 Mesa 0.250 M X 18.3ml MHA 3.44 9 751.9 MAA 0.004575 mon HA is 1752) 9/mo Aug! molar man of (6) At equina lence tonal 25 ml + 18.3 m L olume 2 43.3 mu and mokes of Á =) 0.004575 mon CS Scanned with CamScanner

0.004575 mol and [A-] 0.10566 M 0.0433 L OHT + HA A + Ho 0 0 tu 0.00566 tu n 0.10566-n 2.00 OHS 12.00, P since pH = 0.01 M 2 X 10 soi Сон-) Bare + Aud pkq long Р C0.10566-0.01) + pką lng 10.01) 12.00 11.02 0.98 12.00 pka 11.0 So, fue pką CS Scanned with CamScanner

(6) original pH of the soph =) (HAJI initial 3.56 g 752% 0.183M 0.025 L -11 kq 11.0 + + pe o HA + 0 0 th L 0.183 +n с -n n E 6.183-n x-x [^] [ Monte] 0.183-2 [HA] 6 1:35 X10 1.0x10" 0:19 1-35 tiond My e Now [bot] Emot Log (1.35*10%) pH -log H 5. 87 Р CS Scanned with CamScanner

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