Please write out each step of solving these problems in detail withthe solution.

Question

Please write out each step of solving these problems in detail withthe solution..I really need to understand what is happening herethroughout the process.....Thank you

1) Oxalic acid is a diprotic acid. Calculate thepercent of oxalic acid (H2 C2O4) in a solid given that a 0.7984 g sampleof that solid required 37.98 mL of 0.2283 M NaOH forneutralization.

2) A 25.5 mL aliquot of HCL (aq) of unknown concentration wastitrated with 0.113 M NaOH (aq). It took 51.2 mL of the baseto reach the endpoint of the titration. What was theconcentration (M) of the acid?

Explanation / Answer

M1=MOLARITY OF ACID V1=VOLUME OF ACID
M2=MOLARITY OF BASE V2=VOLUME OF BASE B) 22.5*M1=.113*51.2 M1=0.257M A) MOLES OF NAOH IN 37.98 ML IS=0.0086 MOLES AS OSCALIC ACID IS DIPROTIC SO 2 MOLES OF NAOH USES 1 MOLE OF ACID HENCE ACID USED=0.0043 MOLES=.387 GRAM PECENTAGE=48.4% M2=MOLARITY OF BASE V2=VOLUME OF BASE B) 22.5*M1=.113*51.2 M1=0.257M A) MOLES OF NAOH IN 37.98 ML IS=0.0086 MOLES AS OSCALIC ACID IS DIPROTIC SO 2 MOLES OF NAOH USES 1 MOLE OF ACID HENCE ACID USED=0.0043 MOLES=.387 GRAM PECENTAGE=48.4% M1V1=M2V2 M1=MOLARITY OF ACID V1=VOLUME OF ACID
M2=MOLARITY OF BASE V2=VOLUME OF BASE B) 22.5*M1=.113*51.2 M1=0.257M A) MOLES OF NAOH IN 37.98 ML IS=0.0086 MOLES AS OSCALIC ACID IS DIPROTIC SO 2 MOLES OF NAOH USES 1 MOLE OF ACID HENCE ACID USED=0.0043 MOLES=.387 GRAM PECENTAGE=48.4%

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