M base X M1 unknown V HCl V2 250 mL 00250L M HCl M2 010M M1V1M2V2 sf2 M1 00502L

# M base x m1 unknown v hcl v2 250 ml 00250l m hcl m2

This preview shows page 7 - 9 out of 9 pages.

M (base X) = M1 = unknown V (HCl) = V2 = 25.0 mL = 0.0250L M (HCl) = M2 = 0.10M M1V1=M2V2 sf=2 M1 (0.0502L Base X) = (0.1M HCl)(0.0250L) M1=0.04980=0.050M Base X => The molarity of Base X based on our procedure is 0.050M VI. Analysis & Discussion Questions: 1. Calculate the volume of 0.125 M NaOH needed to neutralize 25.00 mL of 0.100 M HCl. Include a balanced chemical equation in your answer. NaOH(aq) + HCl(aq) = H 2 O(l) + NaCl(aq) sf=3 M1V1=M2V2 (25.00mL HCl)(0.100M)=V2 (0.125M NaOH) V2 = (2.50mol HCl) /(0.125M NaOH) = 20.0mL => 20.0 mL of 0.125M NaOH is needed to neutralize 25.00mL of 0.100M HCl 2. Discuss the titration process - was your titration successful? What was your percentage error for both unknown solutions? (Obtain the actual concentrations from your teacher) Percentage Error(Actual concentration - Theoretical concentration)/Theoretical concentration)*100 Acid R: ((|0.20M - 0.0500M|)/0.0500M)*100%%=3.0*10^2 % (sf=2) Acid Q: ((|0.014M - 0.012M|)/0.012M)*100% = 16.667=17% (sf=2)   #### You've reached the end of your free preview.

Want to read all 9 pages?

• Fall '15
• Neethling,Melanie
• Chemistry, ml, Sodium hydroxide
• • •  