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having trouble balancing the equation in the beginning of this problem. What volume in mL of 0.200 M NaOH do we need to titrate 35.00 mL of 0.140 M HBr to the equivalent point?
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I don't know how to work the problem..but this is what i got for the balancing part NaOH + HBr = H2O + NaBr Everything seems to be already balanced there is no need for coefficients
okay thank you!
NaOH + HBr ---> NaBr + H2O Moles of HBr: .035 * 0.140 M = .0049 mol Volume of NaOH : .0049 / .2 = .0245L = 24.5 ml
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