Question

I need help answering the following chemistry lab questions

Pre-Lab Questions 1. Define the following terms: a) Acid b) Analyte c) Base d) End Point e) Neutralization reactiorn f) Standard solution 8) Titrant h) Titration A 25.00 ml. sample of NaOH was titrated with a 0.743 M HSO, Solution. The endpoint of the titration was observed after the addition of 5.00 ml. of HSO. Calculate the concentration of the NaOH 2. solution. Post-Lab Questions Calculate the volume of a 0.0321 M NaOll that will be required to neutralize 25.00 mL of a 0.0399 M 1. hydrochloric acid solution. Calculate the molarity of the HCE solutions if 25.00 ml. of the acid are required to neutralize 1.00 g of 2. each of following substances: a) NaOH b) Cа(ОН): 61

Answer 1

Aci d preton CH) chamical eadons sabstauce uich Jonte An act is a an elactron pi in the accapt Anal-ta Sulstante shese nshtoprope tkes ane to Le ideanti fied and weasued In Simple, an aalyte solubon of unenaon Goncatahon Base d ion (o) A e is Sste wichdonte a acet a ptnc) o Aeachons dents electzon paiin the chenical dun Ed pint tithn, tdick the indiah chaes a ay t En pint Meles of Males of C Nofmoles ofHt bse acid Neutyalianjeachon base topnduce Salt oetuean an ad ad echion ১ - ১ neutlization eachon Sat t waty haat of neutaali- knouon a AadBase standand selution he solution shase Canceutyhan is kno called standand Solton

G Tt te ihat is oluinof kman Contetsati unay ken Luette and is wsed to detumine thaConutahion n a an unknon sohon Tihehon ateimenh pecedute in ich SolutionConCtinis lemen isslolyr added to a anothe Saluhn to ditmine it aneatahion tendand A tahin is a an kmousrn Vdame 24 Newhalizatin paip VaMa 4Mb na u aVVelumes of acid and se used in netlizalin MH Gnceuthahen of acid and bnse usd in nautelizahn bse obtaine fom the the ntutliatin whee nakn Mdo of acid and balanced chemital for ,Ma743 H ,V= S.oom nE -1, =25 0mL, M S.0omLxo43 M 25.00MLM 2 x o.743 M Soo 2 25.00x MbE 743 M o.2972 M 2S.oo ConCantahn f Nas Sluhen .2972M

Nocl + 2o NaoH Hl Mo 032IH, V?, 25.00mL,M=o.0317M Heu na, n Oo299 M x 2S-oomL o.0221M xV na Иь 0-0299M25.o0m o.97S mL o-032 o221 31.0mL Volune Neo 31 07 m AtNatalizahn,M Hed+ NaoH Nhel + H2o Heie nal,1 (:) le of se No-of est add Ceven M V2500ML 2500 L TOo0 Mass of se Anse MVeNofale 21997 Mlar Mas Mx 25.00 0025 mol ooo o.025 mol x lSDO 2500 L M Moanit MEVb Max Va Het Cae nas2, n Cact, t2o ay.013l Mlo mase of tmse Mx 2500 L = 0.0135 ol o.0135mol x loo 2SiooL =0.S4 M Ma