Question

A student used 0.29 g of a sample and added to it 50.0 mL of 0.100 M HCl. The resulting solution then required 39.50 mL of 0.0980 M NaOH to reach a pH of 3.0. What is the acid consuming capacity of the sample in moles of acid neutralized per gram of sample?

Answer 1

`0.15(mol)/(g)`

Step-by-step explanation:

The first step is to calculate the total amount of moles of HCl, so:

`500 mL = 0.5 L\\\\M=(mol)/(L) \\\\\\mol=0.5L*0.1M=0.05 mol~ HCl`

Then we can calculate the moles of HCl consumed in the reaction of NaOH:

`HCl~ +~ NaOH~ ->~ NaCl~ +~ H_2O`

`39.5~mL=~0.0395~L\\\\mol~=~0.0395L*0.098M=~0.0038~mol~NaOH\\\\0.0038~mol~NaOH=~0.0038~mol~HCl`

Then we have to substract from the intial value to obtain the moles of HCl that react with the sample:

`0.05-0.0038=0.0461~mol~HCl`

Finally to know the ratio between grams and moles we have to divide:

`(0.0461~mol)/(0.29~g)`

`0.15(mol)/(g)`