Calculate how much 0.100 M HCl is needed to neutralize completely 20.0 ml of 0.545 M KOH?

Answer 1

Okay, first of all, the "N" means normality. HBr is hydrobromic acid, which is monoprotic, meaning there's just 1 hydrogen. This makes things easier because molarity (M) = normalilty (N) in monoprotic acids. Assuming you're asking about it neutralizing 0.1 molar KOH, here's how you do it:

Set up a direct proportion and don't forget to convert those mL to liters!

0.20M/x=0.10/0.02345

Solve for x to get: 0.0469 liters, which is 4.69 mL.

Answer 2

Simple equality.

(X volume)(0.200 M HCl) = (10.0 mL)(0.400 M KOH)

0.200X = 4

X = 20 milliliters hydrochloric acid needed

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Answer 3

KOH + HCl -> KCl + H2O

(one to one to completely react)

(0.545 M KOH * 20mL)/0.1 M = mL HCl (0.1M) <109 mL HCl>

<Shorthand>

Answer 4

100mL is needed to neutralize 50 ml of a 0.2 molar solution of HCL.

Answer 5

.20M HClO x .1L= 0.02 moles of HClO

0.3M KClO x .2L= 0.06 moles of KClO

you then use the equation ph= pka + log(KClO/HClO)

ph= 7.538 + 0.477= 8.015