Question

I can't seem to figure out how to set up charts so there's just a screenshot of the question in the file attached :) Thank's for any help!!

Answer 1

See Explanation

Step-by-step explanation:

Volume of base used = (31.79 - 1.6) + (32.12 - 0.98) + (42.87 -13.21)/3 = 30.34 cm3

Volume of acid used = (24.86 - 0.42) + (24.11 - 0.11) + (27.67 -3.42)/3 = 24.23 cm3

For trial 1= number of moles of HBr = 24.44/1000 * 0.3 = 0.0073 moles

For trial 2= number of moles of HBr = 24/1000 * 0.3 = 0.0072 moles

For trial 3= number of moles of HBr = 0.0089 moles

HBr(aq) + KOH(aq) -----> KBr(aq) + H2O(l)

One mole of base is required to neutralize one mole of acid.

From CAVA/CBVB =nA/nB

CAVAnB = CBVBnA

CB = CAVAnB/VBnA

CB = 0.3 * 24.23 * 1/30.34 * 1

CB = 0.24 moldm-3

number of moles of KOH = concentration * volume

number of moles = mass/molar mass

molar mass of KOH = 56 g/mol

mass/56 = 0.24 * 30.34/1000

mass = 0.24 * 30.34/1000 * 56

mass =0.408 g of KOH