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#1
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I am trying to convert from K-lyte potassium tablets which are 25mEq to another potassium tablet which has 40 milligrams per tablet in order to know how much of the tablet to take. please help. thank you
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#2
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Quote:
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#3
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You cannot correctly calculate the equivalent amount of your potassium tablet unless you know which potassium the tablet is. For example: Potassium gluconate contains 4.3mEq per 1000mg, Potassium citrate contains 9.8mEq per 1000mg, Potassium Bicarbonate contains 10mEq per 1000mg, Potassium Acetate contains 10.2mEq per 1000mg and Potassium Chloride contains 13.4mEq per 1000mg. K-lyte is Potassium Chloride. Note also that elemental potassium is 39mg per 1mEq, but potassium chloride is 75mg per 1mEq. Confused? join the crowd.
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#4
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[pottasium gluconate coversion mEq to mg
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#5
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maybe you can, since converting mEq to mg is done by multiplying mEq by its milliequivalent weight, so dividing the mg value of potassium in the other pill which is 40 mg by its (potassium's) milliequivalent weight which is 39.098, you'll get 24.43625 tablets, that's a lot of tablets! anyway I may be wrong, but another calculation seems to check it, using your 978 mg value: 978 mg/40 mg of the replacement pill= 24.45 tablets!
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#6
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[QUOTE=Unregistered;68269][pottasium gluconate coversion mEq to mg[/QU
I am trying to calculate gluconate mEq/L The formulation contains 19 g calcium gluconate/100 mL and 4.5 g calcium glucoheptonate/100 mL Could you please help me Thank you for your assistance |
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#7
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[QUOTE=Unregistered;72910]
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"Equivalents" and "Normality" used to be used to describe how much of a solution would react with something else. You had to find how many electrons per molecule were taking part in the reaction, and calculate the electronic (or equivalent) concentration of the solution based on that. a 1N solution had a concentration of 1 equivalent per litre. What is being discussed here seems to be something quite different. If anyone cares to enlighten us, that would be very nice.
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#8
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[QUOTE=Unregistered;72910]
Quote:
I'm a little confused too, but I think the first line is a quote from an earlier problem and irrelevant to the question. There is no tablet. (I think) Unregistered, So you have 190 g/L of calcium gluconate and 45 g/L of calcium glucoheptonate. Per Wikipedia, the molar masses are 430.373 g/mol and 490.425 g/mol respectively. 190 g/L x 1 mol/430.373 g = 441.5 mmol/L of calcium gluconate. That ionizes to 441.5 mmol of Ca++, hence 883 mEq/L of calcium, and 883 mmol of (C6H11O7)-, or 883 mEq/L (You have additional calcium and glucoheptonate ions present from the other compound, but you only asked about gluconate.) |
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#9
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In that case, the normality of the solution might depend on what is happening with the gluconate. The number of electrons per molecule to balance the equation will depend whether it is going to an acid or CO2 (for example).
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#10
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25 mEq k : 975 mg k per tablet
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