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How many grams of KOH are needed to neutralize 11.5 ml of 0.14 m HCl in stomach acid?

Accepted Answer

To neutralize 11.5 mL of 0.14 M HCl, we need to calculate the number of moles of HCl present:

n = C * V
where,
C = concentration in mol/L = 0.14 M
V = volume in L = 11.5 mL = 0.0115 L

Hence, the number of moles of HCl is:

n = 0.14 M * 0.0115 L
= 0.00161 mol HCl

According to the chemical equation:

HCl + KOH ---→ KCl + H2O

1 mol HCl reacts with 1 mol KOH

Therefore, the number of moles of KOH needed is also 0.00161 mol.

Now, we can convert moles of KOH to grams using its molar mass (56.1 g/mol):

mass = n * molar mass
= 0.00161 mol * 56.1 g/mol
= 0.0903 g KOH

Hence, 0.0903 g of KOH is needed to neutralize 11.5 mL of 0.14 M HCl in stomach acid.

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