A student found that 14.34 ml of 0.5000M HCl was required to titrate all the bor
ID: 75199 • Letter: A
Question
A student found that 14.34 ml of 0.5000M HClwas required to titrate all the borate at one
particular temperature. Assuming this person
used the procedure (5.0 ml sample) described in the manual,
please calculate:
a) number of moles of H+ (H3O+) used
b) number of moles of borate present
c) borate concentration in moles/liter
d) the K value for the process A student found that 14.34 ml of 0.5000M HCl
was required to titrate all the borate at one
particular temperature. Assuming this person
used the procedure (5.0 ml sample) described in the manual,
please calculate:
a) number of moles of H+ (H3O+) used
b) number of moles of borate present
c) borate concentration in moles/liter
d) the K value for the process
Explanation / Answer
The stoicheiometric equation between the given acid andborate is (Let us assume that borate is a sodium borate) Na2B4O7 + 2HCl+5H2O ---------------------------->4H3PO3 + 2NaCl Hence according to equation 1 mole of sodium boratereacts with 2 moles of HCl Therefore a) The number of moles of HCl used in the reaction =Molarity*volume = 0.01434 L*0.5 mol/L = 0.00717 mol b) According above equation The number of moles of borate should be used in thereaction = (0.00717 mol)*(1 mol Borate / 2 mole HCl) = 0.003585 mol c) Total volume in the solution = 14.34 mL + 5.0mL = 19.34 mL Therefore concentration of the borate solution = numberof moles / volume(in L) = 0.003585 mol / 0.01934 L =0.185 MRelated Questions
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